Sine And Cosine Formulae And Their Applications

Let a, b, c be the sides of the given triangle. Then by applying the sine rule, we get

Now substituting the given values we get

(∵ sin (a + b) = sin a cos b + sin b cos a)

Now substituting the corresponding values, we get,

Multiplying 2√2, we get

Hence the ratio of the sides of the given triangle is

We know in a triangle,

∠A + ∠B + ∠C = 180°

⇒ ∠A = 180° - ∠B - ∠C

Substituting the given values, we get

∠A = 180° - 45° - 105°

⇒ ∠A = 30°

Let a, b, c be the sides of the given triangle. Then by applying the sine rule, we get

Now substituting the corresponding values we get

Substitute the equivalent values of the sine, we get

Hence the value of b is 2√2 units.

Let a, b, c be the sides of the given triangle. Then by applying the sine rule, we get

Now substituting the given values we get

Similarly,

Now substituting the given values we get

And given ∠C = 90°, so sin C = sin 90° = 1.

Hence the values of sin A, sin B, sin C are respectively

Let a, b, c be the sides of any triangle ABC. Then by applying the sine rule, we get

⇒a = k sin A

Similarly, b = k sin B

So, a - b = k(sin A - sin B)

And a + b = k(sin A + sin B)

So, the given LHS becomes,

But,

Substituting the above values in equation (i), we get

Rearranging the above equation we get,

Hence proved

Let a, b, c be the sides of any triangle ABC. Then by applying the sine rule, we get

⇒a = k sin A

Similarly, b = k sin B

So, a - b = k(sin A - sin B)..(i)

So the given LHS becomes,

Substituting equation (i) in above equation, we get

But,

Substituting the above values in equation (ii), we get

Rearranging the above equation we get

But

So the above equation becomes,

But from sine rule,

So the above equation becomes,

Hence proved

Let a, b, c be the sides of any triangle ABC. Then by applying the sine rule, we get

⇒a = k sin A

Similarly, b = k sin B

And c = k sin C…..(i)

So, a - b = k(sin A - sin B)..(ii)

So the given LHS becomes,

Substituting equation (i) and (ii) in the above equation, we get

Applying half angle rule,

And

Substituting equation (iii) and (iv) in equation (ii), we get

But

And

So the above equations in equation (v), we get

Dividing numerator and denominator by, we get

By canceling the like terms we get,

Hence proved

Let a, b, c be the sides of any triangle ABC. Then by applying the sine rule, we get

⇒a = k sin A

Similarly, b = k sin B

And c = k sin C…..(i)

So, a + b = k(sin A + sin B)..(ii)

So the given LHS becomes,

Substituting equation (i) and (ii) in the above equation, we get

Applying half angle rule,

And

Substituting equation (iii) and (iv) in equation (ii), we get

But

And

So the above equations in equation (v), we get

Dividing numerator and denominator by, we get

By canceling the like terms we get

Hence proved

Let a, b, c be the sides of any triangle ABC. Then by applying the sine rule, we get

⇒a = k sin A

Similarly, b = k sin B

And c = k sin C…..(i)

So, a + b = k(sin A + sin B)..(ii)

So the given LHS becomes,

Substituting equation (i) and (ii) in above equation, we get

Applying half angle rule,

And

Substituting equation (iii) and (iv) in equation (ii), we get

Hence proved

Let a, b, c be the sides of any triangle ABC. Then by applying the sine rule, we get

⇒c = k sin C

Similarly, b = k sin B

So, b - c = k(sin B - sin C)..(i)

Here we will consider RHS, so we get

Substituting equation (i) in the above equation, we get

But,

Substituting the above values in equation (ii), we get

Rearranging the above equation we get

But

So the above equation becomes,

Hence proved

Let a, b, c be the sides of any triangle ABC. Then by applying the sine rule, we get

⇒c = k sin C

Similarly, b = k sin B

And a = k sin A

Here we will consider LHS, so we get

Substituting corresponding values in the above equation, we get

But,

Substituting the above values in equation (ii), we get

Hence proved

Let a, b, c be the sides of any triangle ABC. Then by applying the sine rule, we get

⇒c = k sin C

Similarly, b = k sin B

And a = k sin A

Here we will consider LHS, so we get

LHS = b sin B – c sin C

Substituting corresponding values in the above equation, we get

⇒ = k sin B sin B –k sin C sin C

⇒ = k (sin² B – sin² C )……….(ii)

But,

Substituting the above values in equation (ii), we get

⇒ = k(sin(B + C) sin(B - C))

But A + B + C = π⇒ B + C = π –A, so the above equation becomes,

⇒ = k(sin(π –A) sin(B - C))

But sin (π - θ) = sin θ

⇒ = k(sin(A) sin(B - C))

From sine rule, a = k sin A, so the above equation becomes,

⇒ = a sin(B - C) = RHS

Hence proved

Let a, b, c be the sides of any triangle ABC. Then by applying the sine rule, we get

⇒c = k sin C

Similarly, b = k sin B

And a = k sin A

Here we will consider RHS, so we get

RHS = (b² – c²) sin A

Substituting corresponding values in the above equation, we get

⇒ = [( k sin B)² - ( k sin C)²] sin A

⇒ = k²(sin² B – sin² C )sin A……….(ii)

But,

Substituting the above values in equation (ii), we get

⇒ = k²(sin(B + C) sin(B - C)) sin A

But A + B + C = π ⇒ B + C = π –A, so the above equation becomes,

⇒ = k²(sin(π –A) sin(B - C))sin A

But sin (π - θ) = sin θ

⇒ = k²(sin(A) sin(B - C))sin A

Rearranging the above equation we get

⇒ = (k sin(A))( sin(B - C))(k sin A)

From sine rule, a = k sin A, so the above equation becomes,

⇒ = a² sin(B - C) = RHS

Hence proved

Let a, b, c be the sides of any triangle ABC. Then by applying the sine rule, we get

Here we will consider LHS, so we get

Multiply and divide by , we get

Substituting corresponding values from sine rule in the above equation, we get

Hence proved

Let a, b, c be the sides of any triangle ABC. Then by applying the sine rule, we get

⇒ a = k sin A, b = k sin B, c = k sin C

Here we will consider LHS, so we get

LHS = a(sin B – sin C) + b(sin C – sin A) + c(sin A – sin B)

Substituting corresponding values from sine rule in above equation, we get

⇒ = k sin A(sin B – sin C) + k sin B(sin C – sin A) + k sin C(sin A – sin B)

⇒ = k sin A sin B – k sin A sin C + k sin B sin C – k sin B sin A + k sin C sin A – k sin C sin B

Cancelling the like terms, we get

LHS = 0 = RHS

Hence proved

Let a, b, c be the sides of any triangle ABC. Then by applying the sine rule, we get

⇒ a = k sin A, b = k sin B, c = k sin C

Here we will consider LHS, so we get

Substituting corresponding values from sine rule in the above equation, we get

Canceling the like terms, we get

But sin(A - B) = sin A cos B – cos A sin B, so the above equation becomes

Cancelling the like terms, we get,

LHS = 0 = RHS

Hence proved

Let a, b, c be the sides of any triangle ABC. Then by applying the sine rule, we get

⇒ a = k sin A, b = k sin B, c = k sin C

Here we will consider LHS, so we get

LHS = a² (cos² B – cos² C) + b² (cos² C – cos² A) + c² (cos² A – cos² B)

Substituting corresponding values from sine rule in above equation, we get

= (k sin A)² (cos² B – cos² C) + (k sin B)² (cos² C – cos² A) + (k sin C)² (cos² A – cos² B)

= k²(sin²A cos² B – sin²A cos² C + sin²B cos² C – sin²B cos² A + sin²A cos² A – sin²A cos² B)

Cancelling the like terms, we get

LHS = 0 = RHS

Hence proved

Let a, b, c be the sides of any triangle ABC. Then by applying the sine rule, we get

⇒ a = k sin A, b = k sin B, c = k sin C

Here we will consider LHS, so we get

LHS = b cos B + c cos C

Substituting corresponding values from sine rule in above equation, we get

⇒ = k sin B cos B + k sin C cos C

Now we will consider RHS, so we get

RHS = a cos (B – C)

Substituting corresponding values from sine rule in the above equation, we get

⇒ = k sin A cos (B – C)

But sin(A + B) + sin (A - B) = 2 sin A cos B, so the above equation becomes,

We know in a triangle, π = A + B + C, hence the above equation becomes,

Comparing equation (i) and (ii),

LHS = RHS

Hence proved

Let a, b, c be the sides of any triangle ABC. Then by applying the sine rule, we get

⇒ a = k sin A, b = k sin B, c = k sin C

Consider the LHS of the given equation, we get

Substituting the values from sine rule into the above equation, we get

Hence proved

Let a, b, c be the sides of any triangle ABC. Then by applying the sine rule, we get

⇒ a = k sin A, b = k sin B, c = k sin C

Now,

Substituting the values from sine rule into the above equation, we get

But

And

And

Substituting these we get

By canceling the like terms, we get

Similarly,

Substituting the values from sine rule into the above equation, we get

But

And

And

Substituting these we get

By canceling the like terms, we get

Similarly,

Substituting the values from sine rule into the above equation, we get

But

And

And

Substituting these we get

By canceling the like terms, we get

So the LHS of the given equation, we get

From equation (i), (ii) and (iii), we get

Hence proved

Let a, b, c be the sides of any triangle ABC. Then by applying the sine rule, we get

⇒ a = k sin A, b = k sin B, c = k sin C

So the LHS of the given equation, we get

Substituting values from sine rule, we get

As A + B + C = π

Hence,

Similarly,

And,

Now , so the above equation becomes,

Canceling the like terms we get

Hence proved

Let a, b, c be the sides of any triangle ABC. Then by applying the sine rule, we get

⇒ a = k sin A, b = k sin B, c = k sin C

So the LHS of the given equation, we get

Substituting values from sine law, we get

Now consider the second part of the equation, we get

Substituting values from sine law, we get

Now consider the third part of the equation, we get

Substituting values from sine law, we get

From equation (i), (ii), and (iii), we get

Hence proved

Let a, b, c be the sides of any triangle ABC. Then by applying the sine rule, we get

⇒ a = k sin A, b = k sin B, c = k sin C

So the LHS of the given equation, we get

LHS = a cos A + b cos B + c cos C

Substituting values from sine law, we get

= k sin A cos A + k sin B cos B + k sin C cos C

(∵ π = A + B + C)

Now, from sine rule,

k sin C = c

Putting this value in equation (i), we get

LHS = 2c sin A sin B

And also k sin B = b (from sine rule)

Putting this in equation (i), we get

LHS = 2b sin A sin C

Hence LHS = RHS

i.e., a cos A + b cos B + c cos C = 2b sin A sin C = 2c sin A sin B

Hence proved

Let a, b, c be the sides of any triangle ABC. Then by applying the sine rule, we get

⇒ a = k sin A, b = k sin B, c = k sin C

So by considering the LHS of the given equation, we get

a(cos B cos C + cos A)

Now substituting the values from sine rule, we get

⇒ = k sin A(cos B cos C + cos A)

⇒ = k (sin A cos B cos C + sin A cos A)

Now consider the second part from the given equation, we get

b(cos A cos C + cos B)

Now substituting the values from sine rule, we get

⇒ = k sin B(cos A cos C + cos B)

⇒ = k (sin B cos A cos C + sin B cos B)

Now consider the third part from the given equation, we get

c(cos A cos B + cos C)

Now substituting the values from sine rule, we get

⇒ = k sin C(cos A cos B + cos C)

⇒ = k (sin C cos A cos B + sin C cos C)

From equation (i), (ii), and (iii), we get

a(cos B cos C + cos A) = b(cos C cos A + cos B) = c(cos A cos B + cos C)

Hence proved

Let a, b, c be the sides of any triangle ABC. Then by applying the sine rule, we get

⇒ a = k sin A, b = k sin B, c = k sin C

So by considering the LHS of the given equation, we get

Substituting the corresponding values from sine rule, we get

Rearranging we get

Regrouping this we get

But

Hence the above equation becomes,

(by applying sine rule)

Hence proved

Let a, b, c be the sides of any triangle ABC. Then by applying the sine rule, we get

⇒ a = k sin A, b = k sin B, c = k sin C

So by considering the RHS of the given equation, we get

RHS = c cos(A - θ) + a cos(C + θ)

Substituting the corresponding values from sine rule, we get

⇒ = k sin C cos(A - θ) + k sin A cos(C + θ)

By cancelling like terms we get

⇒ = k sin B cos θ

⇒ = b cos θ (from sine rule b = k sin B)

= LHS

Hence proved

Let a, b, c be the sides of any triangle ABC. Then by applying the sine rule, we get

So by considering the given condition, we get

sin² A + sin² B = sin² C

Substituting the values from equation (i), we get

This is Pythagoras theorem; hence the given triangle ABC is right - angles triangle

Hence proved

Let a, b, c be the sides of any triangle ABC. Then by applying the sine rule, we get

So by considering the given condition, we get

a², b², c² are in A.P

Then

b² - a² = c² - b² (this is the condition for A.P)

Substituting the values from equation (i), we get

⇒ (k sin B)² - (k sin A)² = (k sin C)² - (k sin B)²

⇒ k² (sin² B - sin² A) = k² (sin² C - sin² B)

⇒ sin (B + A) sin (B - A) = sin (C + B) sin (C - B)

(∵ sin²A - sin²B = sin (A + B) sin (A - B))

⇒ sin (π - C) sin (B - A) = sin (π - A) sin (C - B) (∵ π = A + B + C)

⇒ sin (C) sin (B - A) = sin (A) sin (C - B) (∵ sin (π - θ) = sin θ )

Shuffling this, we get

Canceling the like terms we get

But , so the above equation becomes,

⇒ cot A - cot B = cot B - cot C

Hence cot A, cot B, cot C are in AP

Hence proved

Let BD be the tree, let A be the point where the tree is broken by the wind.

And according to the given condition,

AD = AC

Now let AB = x and AC = y = AD

So the total height of the tree is AB + AD = x + y

Now in ΔABC,

∠C = 30°, ∠B = 90°,

So, ∠A = 180° - (∠B + ∠C) (∵ π = A + B + C)

Hence ∠A = 180° - (90° + 30°) = 60°

Now applying sine rule, we get

Now substituting the values obtained, we get

Substituting the corresponding values, we get

So,

And also from (i),

⇒y = 10√3 m

So the height of the tree is

x + y = 5√3 + 10√3 = 15√3 m

Let AB be the mountain, so the at the foot of a mountain the elevation of its summit is 45^o

So, ∠ACB = 45°

Now when moving on the slope of 30° by a distance of 1000m,

i.e., the CD is the distance moved on the slope of 30° towards the mountain,

Hence CD = 1000m………..(i)

And ∠DCF = 30°

Let EB = FD = x……..(ii)

DE = FB = z…….(iii)

CF = y and AE = t………(iv)

So after moving 1000m, the elevation becomes 60°,

So ∠ADE = 60°

In ΔDFC,

And

Hence

In ΔADE,

⇒t = z√3………(vi)

In ΔABC,

⇒t + x = y + z

⇒z√3 + 500 = 500√3 + z (from (iv), (v), (vi))

⇒z√3-z = 500√3-500

⇒z(√3-1) = 500(√3-1)

⇒z = 500m……(vii)

Hence the equation (vi) becomes,

t = z√3 = (500) √3 m

Hence the height of the mountain is

AB = AE + EB = t + x = (500√3 + 500)m

So the height of the mountain is 500(√3 + 1)m.

Let AB be the peak of a hill, so the at the station A the elevation of its summit is α°

So, ∠CAB = α°

Now when moving on the slope of β° by a distance of ‘c’m,

i.e., AD is the distance moved on the slope of β° towards the hill,

Hence AD = ’c’m………..(i)

And ∠DAF = β°

Let EB = FD = x……..(ii)

DE = FB = z…….(iii)

AF = y and CE = t………(iv)

So after moving ‘c’m, the elevation becomes γ°,

So ∠CDE = γ°

In ΔDFA,

And

In ΔCDE,

⇒z = t cot γ………(vi)

In ΔCBA,

⇒tan α (c cos β + t cot γ ) = t + c sin β

⇒(c tan α cos β + t tan α cot γ ) = t + c sin β

⇒t-t tan α cot γ = c tan α cos β-c sin β

⇒t(1 - tan α cot γ ) = c(tan α cos β-sin β )

Now,

AB = AE + EB = t + x

So the height of the hill is

Hence proved

As a, b, c is in HP (given)

So, are in AP

Hence

Let a, b, c be the sides of any triangle ABC. Then by applying the sine rule, we get

So, a = k sin A, b = k sin B, c = k sin C…(ii)

Substituting equation (ii) in equation (i), we get

By cross multiplying we get

Now, so above equation becomes,

so the above equation becomes

Divide both sides by , we get

Now canceling the like terms we get

Hence are in AP

Therefore,

are in HP

Hence proved

Note: In any ΔABC we define ‘a’ as length of side opposite to ∠A , ‘b’ as length of side opposite to ∠B and ‘c’ as length of side opposite to ∠C .

Keypoint to solve the problem:

Area of ΔABC = ,where θ is the angle between sides BC and AC , a is the length of BC and b is length of AC

We have,

a = 5 , b = 6 and ∠C = 60°

∴ Area of ΔABC =

=

Note: In any ΔABC we define ‘a’ as length of side opposite to ∠A , ‘b’ as length of side opposite to ∠B and ‘c’ as length of side opposite to ∠C .

Keypoint to solve the problem:

Area of ΔABC = ,where θ is the angle between sides BC and AC , a is the length of BC and b is length of AC

We have,

a = 5 , b = 6 and ∠C = 60°

∴ Area of ΔABC =

=

Note: In any ΔABC we define ‘a’ as length of side opposite to ∠A , ‘b’ as length of side opposite to ∠B and ‘c’ as length of side opposite to ∠C .

Keypoint to solve the problem:

• Area of ΔABC = 0.5(product of any two sides)×( sine of angle between them)

• Idea of cosine formula: Cos A =

Given

∴ Cos A =

∵ Area of ΔABC =

We need to find sin A

As we know that - sin²A = 1 – cos²A {using trigonometric identity}

∴ sin A =

∴ ar(ΔABC) = sq units. …ans

Note: In any ΔABC we define ‘a’ as length of side opposite to ∠A , ‘b’ as length of side opposite to ∠B and ‘c’ as length of side opposite to ∠C .

Keypoint to solve the problem:

• Area of ΔABC = 0.5(product of any two sides)×( sine of angle between them)

• Idea of cosine formula: Cos A =

Given

∴ Cos A =

∵ Area of ΔABC =

We need to find sin A

As we know that - sin²A = 1 – cos²A {using trigonometric identity}

∴ sin A =

∴ ar(ΔABC) = sq units. …ans

Note: In any ΔABC we define ‘a’ as length of side opposite to ∠A , ‘b’ as length of side opposite to ∠B and ‘c’ as length of side opposite to ∠C .

Key point to solve the problem:

Idea of cosine formula in ΔABC

• Cos A =

• Cos B =

• Cos C =

As we have a = 5, b = 6 and c = 8

∴ Cos A =

Cos B =

Cos C =

We have to prove:

8 cos A + 16 cos B + 4 cos C = 17

LHS = 8 cos A + 16 cos B + 4 cos C

Putting the values of cos A, cos B and cos C in LHS

LHS =

LHS ≠ RHS

From cosine expressions we have:

96 cos A = 75 , 80 cos B = 53 and 20 cos C = -1

Adding all we have,

96 cos A + 80 cos B +20 cos C = 75+53-1 = 127

∴ 96 cos A + 80 cos B +20 cos C = 127

Please check it….

Note: In any ΔABC we define ‘a’ as length of side opposite to ∠A , ‘b’ as length of side opposite to ∠B and ‘c’ as length of side opposite to ∠C .

Key point to solve the problem:

Idea of cosine formula in ΔABC

• Cos A =

• Cos B =

• Cos C =

As we have a = 5, b = 6 and c = 8

∴ Cos A =

Cos B =

Cos C =

We have to prove:

8 cos A + 16 cos B + 4 cos C = 17

LHS = 8 cos A + 16 cos B + 4 cos C

Putting the values of cos A, cos B and cos C in LHS

LHS =

LHS ≠ RHS

From cosine expressions we have:

96 cos A = 75 , 80 cos B = 53 and 20 cos C = -1

Adding all we have,

96 cos A + 80 cos B +20 cos C = 75+53-1 = 127

∴ 96 cos A + 80 cos B +20 cos C = 127

Please check it….

Note: In any ΔABC we define ‘a’ as length of side opposite to ∠A , ‘b’ as length of side opposite to ∠B and ‘c’ as length of side opposite to ∠C .

Key point to solve the problem:

Idea of cosine formula in ΔABC

•

Cos A =

• Cos B =

• Cos C =

As we have a = 18, b = 24 and c = 30

∴ Cos A =

Cos B =

Cos C =

Note: In any ΔABC we define ‘a’ as length of side opposite to ∠A , ‘b’ as length of side opposite to ∠B and ‘c’ as length of side opposite to ∠C .

Key point to solve the problem:

Idea of cosine formula in ΔABC

•

Cos A =

• Cos B =

• Cos C =

As we have a = 18, b = 24 and c = 30

∴ Cos A =

Cos B =

Cos C =

Note: In any ΔABC we define ‘a’ as length of side opposite to ∠A , ‘b’ as length of side opposite to ∠B and ‘c’ as length of side opposite to ∠C .

Key point to solve the problem:

Idea of cosine formula in ΔABC

• Cos A =

• Cos B =

• Cos C =

As we have to prove:

b (c cos A – a cos C) = c² – a²

As LHS contain bc cos A and ab cos C which can be obtained from cosine formulae.

∴ From cosine formula we have:

Cos A =

⇒ bc cos A = …..eqn 1

And Cos C =

⇒ ab cos C = ……eqn 2

Subtracting eqn 2 from eqn 1:

bc cos A - ab cos C = -

⇒ bc cos A - ab cos C = c² - a²

∴b(c cos A- a cos C) = c² - a² …proved

Note: In any ΔABC we define ‘a’ as length of side opposite to ∠A , ‘b’ as length of side opposite to ∠B and ‘c’ as length of side opposite to ∠C .

Key point to solve the problem:

Idea of cosine formula in ΔABC

• Cos A =

• Cos B =

• Cos C =

As we have to prove:

b (c cos A – a cos C) = c² – a²

As LHS contain bc cos A and ab cos C which can be obtained from cosine formulae.

∴ From cosine formula we have:

Cos A =

⇒ bc cos A = …..eqn 1

And Cos C =

⇒ ab cos C = ……eqn 2

Subtracting eqn 2 from eqn 1:

bc cos A - ab cos C = -

⇒ bc cos A - ab cos C = c² - a²

∴b(c cos A- a cos C) = c² - a² …proved

Note: In any ΔABC we define ‘a’ as length of side opposite to ∠A , ‘b’ as length of side opposite to ∠B and ‘c’ as length of side opposite to ∠C .

Key point to solve the problem:

Idea of cosine formula in ΔABC

• Cos A =

• Cos B =

• Cos C =

As we have to prove:

c (a cos B – b cos A) = a² – b²

As LHS contain ca cos B and cb cos A which can be obtained from cosine formulae.

∴ From cosine formula we have:

Cos A =

⇒ bc cos A = …..eqn 1

And Cos B =

⇒ ac cos B = ……eqn 2

Subtracting eqn 1 from eqn 2:

ac cos B - bc cos A =

⇒ ac cos B - bc cos A =

∴c (a cos B – b cos A) = a² – b² …proved

Note: In any ΔABC we define ‘a’ as length of side opposite to ∠A , ‘b’ as length of side opposite to ∠B and ‘c’ as length of side opposite to ∠C .

Key point to solve the problem:

Idea of cosine formula in ΔABC

• Cos A =

• Cos B =

• Cos C =

As we have to prove:

c (a cos B – b cos A) = a² – b²

As LHS contain ca cos B and cb cos A which can be obtained from cosine formulae.

∴ From cosine formula we have:

Cos A =

⇒ bc cos A = …..eqn 1

And Cos B =

⇒ ac cos B = ……eqn 2

Subtracting eqn 1 from eqn 2:

ac cos B - bc cos A =

⇒ ac cos B - bc cos A =

∴c (a cos B – b cos A) = a² – b² …proved

Note: In any ΔABC we define ‘a’ as length of side opposite to ∠A , ‘b’ as length of side opposite to ∠B and ‘c’ as length of side opposite to ∠C .

Key point to solve the problem:

Idea of cosine formula in ΔABC

• Cos A =

• Cos B =

• Cos C =

As we have to prove:

2 (bc cos A + ca cos B + ab cos C) = a² + b² + c²

As LHS contain 2ca cos B, 2ab cos C and 2cb cos A ,which can be obtained from cosine formulae.

∴ From cosine formula we have:

Cos A =

⇒ 2bc cos A = …..eqn 1

Cos C =

⇒ 2ab cos C = …eqn 2

And, Cos B =

⇒ 2ac cos B = ……eqn 3

Adding eqn 1,2 and 3:-

2bc cos A + 2ab cos C + 2ac cos B = +

⇒ 2bc cos A + 2ab cos C + 2ac cos B =

⇒ 2(bc cos A + ab cos C + ac cos B) = …proved

Note: In any ΔABC we define ‘a’ as length of side opposite to ∠A , ‘b’ as length of side opposite to ∠B and ‘c’ as length of side opposite to ∠C .

Key point to solve the problem:

Idea of cosine formula in ΔABC

• Cos A =

• Cos B =

• Cos C =

As we have to prove:

2 (bc cos A + ca cos B + ab cos C) = a² + b² + c²

As LHS contain 2ca cos B, 2ab cos C and 2cb cos A ,which can be obtained from cosine formulae.

∴ From cosine formula we have:

Cos A =

⇒ 2bc cos A = …..eqn 1

Cos C =

⇒ 2ab cos C = …eqn 2

And, Cos B =

⇒ 2ac cos B = ……eqn 3

Adding eqn 1,2 and 3:-

2bc cos A + 2ab cos C + 2ac cos B = +

⇒ 2bc cos A + 2ab cos C + 2ac cos B =

⇒ 2(bc cos A + ab cos C + ac cos B) = …proved

Note: In any ΔABC we define ‘a’ as length of side opposite to ∠A , ‘b’ as length of side opposite to ∠B and ‘c’ as length of side opposite to ∠C .

The keypoint to solve the problem:

The idea of cosine formula in ΔABC

• Cos A =

• Cos B =

• Cos C =

The idea of sine formula in ΔABC

•

As we have to prove:

(c² – a² + b²) tan A = (a² – b² + c²) tan B = (b² – c² + a²) tan C

As LHS contain (c² – a² + b²), (a² – b² + c²)and (b² – c² + a²),which shows resemblance with cosine formulae.

∴ From cosine formula we have:

Cos A =

⇒ 2bc cos A =

Multiplying with tan A both sides to get the form desired in proof

2bc cos A tan A =

2bc sin A = …..eqn 1

Cos C =

⇒ 2ab cos C =

Multiplying with tan C both sides to get the form desired in proof

2ab cos C tan C = tan C

2ab sin C = tan C ..…eqn 2

And, Cos B =

⇒ 2ac cos B =

Multiplying with tan B both sides to get the form desired in proof

2ac cos B tan B =

2ac sin B = ……eqn 3

As we are observing that sin terms are being involved so let’s try to use sine formula.

From sine formula we have,

⇒

Multiplying abc to each fraction:-

⇒ bc sin A = ac sin B = ab sin C

⇒ 2bc sin A = 2ac sin B = 2ab sin C

∴ From eqn 1, 2 and 3 we have:

= tan C

Hence, proved.

Note: In any ΔABC we define ‘a’ as length of side opposite to ∠A , ‘b’ as length of side opposite to ∠B and ‘c’ as length of side opposite to ∠C .

The keypoint to solve the problem:

The idea of cosine formula in ΔABC

• Cos A =

• Cos B =

• Cos C =

The idea of sine formula in ΔABC

•

As we have to prove:

(c² – a² + b²) tan A = (a² – b² + c²) tan B = (b² – c² + a²) tan C

As LHS contain (c² – a² + b²), (a² – b² + c²)and (b² – c² + a²),which shows resemblance with cosine formulae.

∴ From cosine formula we have:

Cos A =

⇒ 2bc cos A =

Multiplying with tan A both sides to get the form desired in proof

2bc cos A tan A =

2bc sin A = …..eqn 1

Cos C =

⇒ 2ab cos C =

Multiplying with tan C both sides to get the form desired in proof

2ab cos C tan C = tan C

2ab sin C = tan C ..…eqn 2

And, Cos B =

⇒ 2ac cos B =

Multiplying with tan B both sides to get the form desired in proof

2ac cos B tan B =

2ac sin B = ……eqn 3

As we are observing that sin terms are being involved so let’s try to use sine formula.

From sine formula we have,

⇒

Multiplying abc to each fraction:-

⇒ bc sin A = ac sin B = ab sin C

⇒ 2bc sin A = 2ac sin B = 2ab sin C

∴ From eqn 1, 2 and 3 we have:

= tan C

Hence, proved.

Note: In any ΔABC we define ‘a’ as the length of the side opposite to ∠A, ‘b’ as the length of the side opposite to ∠B and ‘c’ as the length of the side opposite to ∠C.

Key point to solve the problem:

Idea of projection Formula:

• c = a cos B + b cos A

• b = c cos A + a cos C

• a = c cos B + b cos C

As we have to prove:

We can observe that we can get terms c – b cos A and b – c cos A from projection formula

∴ from projection formula we have-

c = a cos B + b cos A

⇒ c – b cos A = a cos B …..eqn 1

Also,

b = c cos A + a cos C

⇒ b – c cos A = a cos C ……eqn 2

Dividing eqn 1 by eqn 2, we have-

⇒ Hence proved.

Note: In any ΔABC we define ‘a’ as the length of the side opposite to ∠A, ‘b’ as the length of the side opposite to ∠B and ‘c’ as the length of the side opposite to ∠C.

Key point to solve the problem:

Idea of projection Formula:

• c = a cos B + b cos A

• b = c cos A + a cos C

• a = c cos B + b cos C

As we have to prove:

We can observe that we can get terms c – b cos A and b – c cos A from projection formula

∴ from projection formula we have-

c = a cos B + b cos A

⇒ c – b cos A = a cos B …..eqn 1

Also,

b = c cos A + a cos C

⇒ b – c cos A = a cos C ……eqn 2

Dividing eqn 1 by eqn 2, we have-

⇒ Hence proved.

Note: In any ΔABC we define ‘a’ as the length of the side opposite to ∠A, ‘b’ as the length of the side opposite to ∠B and ‘c’ as the length of the side opposite to ∠C.

Key point to solve the problem:

Idea of projection Formula:

• c = a cos B + b cos A

• b = c cos A + a cos C

• a = c cos B + b cos C

As we have to prove:

a(cos B + cos C – 1) + b(cos C + cos A –1) + c(cos A + cos B – 1) = 0

We can observe that we all the terms present in equation to be proved are also present in expressions of projection formula ,so we have to apply the formula with slight modification -

∴ from projection formula we have-

c = a cos B + b cos A

⇒ b cos A + a cos B – c = 0 …..eqn 1

Also,

b = c cos A + a cos C

⇒ c cos A + a cos C – b = 0 ……eqn 2

Also,

a = c cos B + b cos C

⇒ c cos B + b cos C – a = 0 …..eqn 3

Adding eqn 1 ,2 and 3 –

We have,

b cos A + a cos B – c + c cos A + a cos C – b + c cos B + b cos C – a = 0

b cos A – b + b cos C + a cos B + a cos C – a + c cos A + c cos B – c = 0

⇒ b(cos A + cos C – 1) + a(cos B + cos C – 1) + c(cos A + cos B -1) = 0

Hence,

a(cos B + cos C – 1) + b(cos C + cos A –1) + c(cos A + cos B – 1) = 0

….proved

Note: In any ΔABC we define ‘a’ as the length of the side opposite to ∠A, ‘b’ as the length of the side opposite to ∠B and ‘c’ as the length of the side opposite to ∠C.

Key point to solve the problem:

Idea of projection Formula:

• c = a cos B + b cos A

• b = c cos A + a cos C

• a = c cos B + b cos C

As we have to prove:

a(cos B + cos C – 1) + b(cos C + cos A –1) + c(cos A + cos B – 1) = 0

We can observe that we all the terms present in equation to be proved are also present in expressions of projection formula ,so we have to apply the formula with slight modification -

∴ from projection formula we have-

c = a cos B + b cos A

⇒ b cos A + a cos B – c = 0 …..eqn 1

Also,

b = c cos A + a cos C

⇒ c cos A + a cos C – b = 0 ……eqn 2

Also,

a = c cos B + b cos C

⇒ c cos B + b cos C – a = 0 …..eqn 3

Adding eqn 1 ,2 and 3 –

We have,

b cos A + a cos B – c + c cos A + a cos C – b + c cos B + b cos C – a = 0

b cos A – b + b cos C + a cos B + a cos C – a + c cos A + c cos B – c = 0

⇒ b(cos A + cos C – 1) + a(cos B + cos C – 1) + c(cos A + cos B -1) = 0

Hence,

a(cos B + cos C – 1) + b(cos C + cos A –1) + c(cos A + cos B – 1) = 0

….proved

Note: In any ΔABC we define ‘a’ as the length of the side opposite to ∠A, ‘b’ as the length of the side opposite to ∠B and ‘c’ as the length of the side opposite to ∠C.

The keypoint to solve the problem:

The idea of sine Formula:

•

As we have to prove:

a cos A + b cos B + c cos C = 2b sin A sin C

We can observe that we all the terms present in the equation to be proved are not showing any resemblance with known formula but the term is RHS side has sine terms, so there is a possibility that sine formula can solve our problem

∴ from sine formula we have-

∴ a = 2k sin A , b = 2k sin B , c = 2k sin C

As,

LHS = a cos A + b cos B + c cos C

= 2k sin A cos A + 2k sin B cos B + 2k sin C cos C

= k(2sin A cos A + 2sin B cos B + 2sin C cos C

LHS = k( sin 2A + sin 2B + sin 2C) {using 2 sin X cos X = sin 2X }

Using transformation formula – sin X + sin Y =

LHS = k ( 2sin(A + B) cos (A – B) + sin 2C)

∵ ∠ A + ∠ B + ∠ C = π

∴ A + B = π – C

∴ LHS = k { 2sin (π – C) cos (A – B) + 2 sin C cos C }

[as sin (π – θ) = sin θ]

LHS = k{ 2 sin C cos (A – B) + 2 sin C cos C }

LHS = 2k sin C { cos (A – B) + cos C }

Using transformation formula – cos X + cos Y =

LHS = 2k sin C { }

LHS = 4k sin C {∵∠ A + ∠ B + ∠ C = π }

LHS = 4k sin C sin B sin A

∵ 2k sin B = b

We have,

LHS = 2b sin A sin C = RHS …..Hence proved.

Note: In any ΔABC we define ‘a’ as the length of the side opposite to ∠A, ‘b’ as the length of the side opposite to ∠B and ‘c’ as the length of the side opposite to ∠C.

The keypoint to solve the problem:

The idea of sine Formula:

•

As we have to prove:

a cos A + b cos B + c cos C = 2b sin A sin C

We can observe that we all the terms present in the equation to be proved are not showing any resemblance with known formula but the term is RHS side has sine terms, so there is a possibility that sine formula can solve our problem

∴ from sine formula we have-

∴ a = 2k sin A , b = 2k sin B , c = 2k sin C

As,

LHS = a cos A + b cos B + c cos C

= 2k sin A cos A + 2k sin B cos B + 2k sin C cos C

= k(2sin A cos A + 2sin B cos B + 2sin C cos C

LHS = k( sin 2A + sin 2B + sin 2C) {using 2 sin X cos X = sin 2X }

Using transformation formula – sin X + sin Y =

LHS = k ( 2sin(A + B) cos (A – B) + sin 2C)

∵ ∠ A + ∠ B + ∠ C = π

∴ A + B = π – C

∴ LHS = k { 2sin (π – C) cos (A – B) + 2 sin C cos C }

[as sin (π – θ) = sin θ]

LHS = k{ 2 sin C cos (A – B) + 2 sin C cos C }

LHS = 2k sin C { cos (A – B) + cos C }

Using transformation formula – cos X + cos Y =

LHS = 2k sin C { }

LHS = 4k sin C {∵∠ A + ∠ B + ∠ C = π }

LHS = 4k sin C sin B sin A

∵ 2k sin B = b

We have,

LHS = 2b sin A sin C = RHS …..Hence proved.

Note: In any ΔABC we define ‘a’ as the length of the side opposite to ∠A, ‘b’ as the length of the side opposite to ∠B and ‘c’ as the length of the side opposite to ∠C.

The keypoint to solve the problem:

The idea of cosine formula in ΔABC

• Cos A =

• Cos B =

• Cos C =

As we have to prove:

The form required to prove contains similar terms as present in cosine formula.

∴ Cosine formula is the perfect tool for solving the problem.

As we see the expression has bc term so we will apply the formula of cosA

As cos A =

⇒ 2bc cos A = b² + c² – a²

We need (b + c )² in our proof so adding 2bc both sides –

∴ 2bc + 2bc cos A = b² + c² +2bc – a²

⇒ 2bc ( 1 + cos A) = (b + c)² - a²

∵ 1 + cos A = 2cos² (A / 2) { using multiple angle formulae }

∴

⇒ …..Hence proved.

Note: In any ΔABC we define ‘a’ as the length of the side opposite to ∠A, ‘b’ as the length of the side opposite to ∠B and ‘c’ as the length of the side opposite to ∠C.

The keypoint to solve the problem:

The idea of cosine formula in ΔABC

• Cos A =

• Cos B =

• Cos C =

As we have to prove:

The form required to prove contains similar terms as present in cosine formula.

∴ Cosine formula is the perfect tool for solving the problem.

As we see the expression has bc term so we will apply the formula of cosA

As cos A =

⇒ 2bc cos A = b² + c² – a²

We need (b + c )² in our proof so adding 2bc both sides –

∴ 2bc + 2bc cos A = b² + c² +2bc – a²

⇒ 2bc ( 1 + cos A) = (b + c)² - a²

∵ 1 + cos A = 2cos² (A / 2) { using multiple angle formulae }

∴

⇒ …..Hence proved.

Note: In any ΔABC we define ‘a’ as the length of the side opposite to ∠A, ‘b’ as the length of the side opposite to ∠B and ‘c’ as the length of the side opposite to ∠C.

The keypoint to solve the problem:

The idea of cosine formula in ΔABC

• Cos A =

• Cos B =

• Cos C =

As we have to prove:

= (a+b+c)²

The form required to prove contains similar terms as present in cosine formula.

∴ Cosine formula is the perfect tool for solving the problem.

As we see the expression has bc, ac and ab terms so we will apply the formula of cos A, cos B, and cos C all.

As cos A =

⇒ 2bc cos A = b² + c² – a²

We need (b + c )² in our proof so adding 2bc both sides –

∴ 2bc + 2bc cos A = b² + c² +2bc – a²

⇒ 2bc ( 1 + cos A) = (b + c)² - a²

∵ 1 + cos A = 2cos² (A / 2) { using multiple angle formulae }

∴

⇒

⇒ ….eqn 1

Similarly,

….eqn 2

And,
….eqn 3

Adding equation 1, 2 and 3 we have –

…Hence proved

Note: In any ΔABC we define ‘a’ as the length of the side opposite to ∠A, ‘b’ as the length of the side opposite to ∠B and ‘c’ as the length of the side opposite to ∠C.

The keypoint to solve the problem:

The idea of cosine formula in ΔABC

• Cos A =

• Cos B =

• Cos C =

As we have to prove:

= (a+b+c)²

The form required to prove contains similar terms as present in cosine formula.

∴ Cosine formula is the perfect tool for solving the problem.

As we see the expression has bc, ac and ab terms so we will apply the formula of cos A, cos B, and cos C all.

As cos A =

⇒ 2bc cos A = b² + c² – a²

We need (b + c )² in our proof so adding 2bc both sides –

∴ 2bc + 2bc cos A = b² + c² +2bc – a²

⇒ 2bc ( 1 + cos A) = (b + c)² - a²

∵ 1 + cos A = 2cos² (A / 2) { using multiple angle formulae }

∴

⇒

⇒ ….eqn 1

Similarly,

….eqn 2

And,
….eqn 3

Adding equation 1, 2 and 3 we have –

…Hence proved

The keypoint to solve the problem:

The idea of sine Formula:

•

Idea of projection Formula:

• c = a cos B + b cos A

• b = c cos A + a cos C

• a = c cos B + b cos C

As we have to prove:-

sin³ A cos (B – C) + sin³ B cos (C – A) + sin³ C cos (A – B) = 3 sin A sin B sin C

as there is no resemblance of above expression with any formula so first we need to simplify the expression

LHS = sin³ A cos (B – C) + sin³ B cos (C – A) + sin³ C cos (A – B)

LHS = sin² A sin A cos (B – C) + sin² B sin B cos (C – A) + sin² C sin C cos (A - B)

LHS = sin² A sin{π – (B+C)}cos (B – C) + sin²B sin{π – (A+C)}cos (C – A) + sin²C sin {π – (A + B)} cos (A - B)

LHS = sin²A sin (B+C) cos(B-C) + sin²B sin(A + C)cos(C – A) + sin²C sin(B + C) cos (A - B)

Using the relation sin ( X + Y )cos(X – Y) = sin 2X + sin 2Y , we have –

LHS = sin²A (sin 2B + sin 2C) + sin²B (sin 2A + sin 2C) + sin²C (sin 2B + sin 2A)

Using sin 2X = 2sin X cos X , we have –

LHS = sin²A (2sinB cosB + 2sinC cosC) + sin²B (2sinA cosA + 2sinC cosC) + sin²C (2sinBcosB + 2sinA cosA)

Using sine formula we have –

∴ sin A = ka , sin B = kb and sin C = kc …eqn 1

Putting the values in LHS:-

LHS =

LHS =

Using projection formula

• c = a cos B + b cos A

• b = c cos A + a cos C

• a = c cos B + b cos C

We have

LHS =

= {using eqn 1}

= = RHS ….Hence proved

The keypoint to solve the problem:

The idea of sine Formula:

•

Idea of projection Formula:

• c = a cos B + b cos A

• b = c cos A + a cos C

• a = c cos B + b cos C

As we have to prove:-

sin³ A cos (B – C) + sin³ B cos (C – A) + sin³ C cos (A – B) = 3 sin A sin B sin C

as there is no resemblance of above expression with any formula so first we need to simplify the expression

LHS = sin³ A cos (B – C) + sin³ B cos (C – A) + sin³ C cos (A – B)

LHS = sin² A sin A cos (B – C) + sin² B sin B cos (C – A) + sin² C sin C cos (A - B)

LHS = sin² A sin{π – (B+C)}cos (B – C) + sin²B sin{π – (A+C)}cos (C – A) + sin²C sin {π – (A + B)} cos (A - B)

LHS = sin²A sin (B+C) cos(B-C) + sin²B sin(A + C)cos(C – A) + sin²C sin(B + C) cos (A - B)

Using the relation sin ( X + Y )cos(X – Y) = sin 2X + sin 2Y , we have –

LHS = sin²A (sin 2B + sin 2C) + sin²B (sin 2A + sin 2C) + sin²C (sin 2B + sin 2A)

Using sin 2X = 2sin X cos X , we have –

LHS = sin²A (2sinB cosB + 2sinC cosC) + sin²B (2sinA cosA + 2sinC cosC) + sin²C (2sinBcosB + 2sinA cosA)

Using sine formula we have –

∴ sin A = ka , sin B = kb and sin C = kc …eqn 1

Putting the values in LHS:-

LHS =

LHS =

Using projection formula

• c = a cos B + b cos A

• b = c cos A + a cos C

• a = c cos B + b cos C

We have

LHS =

= {using eqn 1}

= = RHS ….Hence proved

Note: In any ΔABC we define ‘a’ as the length of the side opposite to ∠A, ‘b’ as the length of the side opposite to ∠B and ‘c’ as the length of the side opposite to ∠C.

The keypoint to solve the problem:

The idea of cosine formula in ΔABC

• Cos A =

• Cos B =

• Cos C =

As we have to prove under given conditions.

∵ Only cos terms are involved so we will apply cosine formula to find cos A , cos B, and cos C and we will take their ratio.

∵

∴ b + c = 12k ….eqn 1

c + a = 13k ….eqn 2

a + b = 15k ….eqn 3

But only above relation is not sufficient to find cosines as k is unknown, either we need to express k in terms of a , b or c or express a , b , c in terms of k. Later part is easier.

∴ we will find a,b,c in terms of k

Adding eqn 1,2 and 3 we have –

2 (a + b + c) = 40k

∴ a + b+ c = 20k

∴ a = 20k – (b + c) = 20k – 12k = 8k

Similarly, b = 20k – (c + a) = 20k – 13k = 7k

And c = 20k – (a + b) = 20k – 15k = 5k

Hence,

Cos A =

Cos B =

cos C =

∴

∴ ….Hence proved.

Note: In any ΔABC we define ‘a’ as the length of the side opposite to ∠A, ‘b’ as the length of the side opposite to ∠B and ‘c’ as the length of the side opposite to ∠C.

The keypoint to solve the problem:

The idea of cosine formula in ΔABC

• Cos A =

• Cos B =

• Cos C =

As we have to prove under given conditions.

∵ Only cos terms are involved so we will apply cosine formula to find cos A , cos B, and cos C and we will take their ratio.

∵

∴ b + c = 12k ….eqn 1

c + a = 13k ….eqn 2

a + b = 15k ….eqn 3

But only above relation is not sufficient to find cosines as k is unknown, either we need to express k in terms of a , b or c or express a , b , c in terms of k. Later part is easier.

∴ we will find a,b,c in terms of k

Adding eqn 1,2 and 3 we have –

2 (a + b + c) = 40k

∴ a + b+ c = 20k

∴ a = 20k – (b + c) = 20k – 12k = 8k

Similarly, b = 20k – (c + a) = 20k – 13k = 7k

And c = 20k – (a + b) = 20k – 15k = 5k

Hence,

Cos A =

Cos B =

cos C =

∴

∴ ….Hence proved.

The keypoint to solve the problem:

The idea of cosine formula in ΔABC

Cos A = Cos B = Cos C = .

As we have to prove : (a + b + c) (a – b + c) = 3ca

LHS = (a + c+ b) (a + c– b) = (a + c)² – b² { using ( x + y )( x – y ) = x² – y² }

Now the above expression gives us hint that we need to apply cosine formula as terms has resemblance.

∴ cos B =

cos 60° =

⇒ =

∴ ac =

Adding 2bc both sides to get the term present in final term-

∴ 3ac = a² + c² + 2ac – b²

⇒ 3ac = (a + c)² – b²

using ( x + y )( x – y ) = x² – y² , we have –

3ac = (a + c+ b) (a + c– b)

Or (a + b + c) (a – b + c) = 3ca …Hence proved

The keypoint to solve the problem:

The idea of cosine formula in ΔABC

Cos A = Cos B = Cos C = .

As we have to prove : (a + b + c) (a – b + c) = 3ca

LHS = (a + c+ b) (a + c– b) = (a + c)² – b² { using ( x + y )( x – y ) = x² – y² }

Now the above expression gives us hint that we need to apply cosine formula as terms has resemblance.

∴ cos B =

cos 60° =

⇒ =

∴ ac =

Adding 2bc both sides to get the term present in final term-

∴ 3ac = a² + c² + 2ac – b²

⇒ 3ac = (a + c)² – b²

using ( x + y )( x – y ) = x² – y² , we have –

3ac = (a + c+ b) (a + c– b)

Or (a + b + c) (a – b + c) = 3ca …Hence proved

The keypoint to solve the problem:

The idea of basic trigonometric formulae, i.e. transformation and T – ratios of multiple angles

Given,

cos²A + cos² B + cos² C = 1

Multiplying 2 to both sides so that we can change it in Trigonometric ratios of multiple angles so that we can get the value of angle.

As 2 cos²X = 1 + cos 2X

∴ 2cos²A + 2cos² B + 2cos² C = 2

⇒ 1 + cos 2A + 1 + cos 2B + 1 + cos 2C = 2

⇒ cos 2A + cos 2B + cos 2C = -1

Using, cos 2X + cos 2Y = 2 cos (X + Y) cos (X – Y)

⇒ 2 cos (A + B) cos (A – B) = -1(1 + cos 2C)

As 2 cos²X = 1 + cos 2X and A + B + C = π

We have,

2 cos (π - C) cos (A – B) = -2 cos² C

-2 cos C cos (A – B) = -2 cos² C {∵ cos (π - θ) = - cos θ }

∴ 2 cos C ( cos C + cos (A – B)) = 0

Either cos C = 0 ⇒ ∠ C = 90°

Or cos C = -cos (A – B) ⇒ C = π – (A – B) which is not possible as in ΔABC, A + B + C = π

∴ C = 90° is the only satisfied solution.

Hence, Δ ABC is a right triangle, right angled at ∠ C …proved

The keypoint to solve the problem:

The idea of basic trigonometric formulae, i.e. transformation and T – ratios of multiple angles

Given,

cos²A + cos² B + cos² C = 1

Multiplying 2 to both sides so that we can change it in Trigonometric ratios of multiple angles so that we can get the value of angle.

As 2 cos²X = 1 + cos 2X

∴ 2cos²A + 2cos² B + 2cos² C = 2

⇒ 1 + cos 2A + 1 + cos 2B + 1 + cos 2C = 2

⇒ cos 2A + cos 2B + cos 2C = -1

Using, cos 2X + cos 2Y = 2 cos (X + Y) cos (X – Y)

⇒ 2 cos (A + B) cos (A – B) = -1(1 + cos 2C)

As 2 cos²X = 1 + cos 2X and A + B + C = π

We have,

2 cos (π - C) cos (A – B) = -2 cos² C

-2 cos C cos (A – B) = -2 cos² C {∵ cos (π - θ) = - cos θ }

∴ 2 cos C ( cos C + cos (A – B)) = 0

Either cos C = 0 ⇒ ∠ C = 90°

Or cos C = -cos (A – B) ⇒ C = π – (A – B) which is not possible as in ΔABC, A + B + C = π

∴ C = 90° is the only satisfied solution.

Hence, Δ ABC is a right triangle, right angled at ∠ C …proved

The keypoint to solve the problem:

To prove a triangle isosceles our task is to show either any two angles equal or two sides equal.

Idea of cosine formula - Cos C =

The idea of sine Formula:

•

Given,

As it has sin terms involved so that sine formula can work, and cos C is also there so we might need cosine formula too.

Let’s apply sine formula keeping a target to prove any two sides equal.

Using sine formula we have –

∴ sin A = ak and sin B = bk

∴ cos C =

If we apply cosine formula, we will get an equation in terms of sides only that may give us any two sides equal.

Using, Cos C =

We have,

⇒ b² + a² – c² = a²

⇒ b² = c²

⇒ b = c

Hence 2 sides are equal.

∴ Δ ABC is isosceles. ….proved

The keypoint to solve the problem:

To prove a triangle isosceles our task is to show either any two angles equal or two sides equal.

Idea of cosine formula - Cos C =

The idea of sine Formula:

•

Given,

As it has sin terms involved so that sine formula can work, and cos C is also there so we might need cosine formula too.

Let’s apply sine formula keeping a target to prove any two sides equal.

Using sine formula we have –

∴ sin A = ak and sin B = bk

∴ cos C =

If we apply cosine formula, we will get an equation in terms of sides only that may give us any two sides equal.

Using, Cos C =

We have,

⇒ b² + a² – c² = a²

⇒ b² = c²

⇒ b = c

Hence 2 sides are equal.

∴ Δ ABC is isosceles. ….proved

The keypoint to solve the problem:

The idea of cosine formula –

Cos C = Cos A = Cos B =

According to the question:

One ship goes in north east direction while other in southeast direction.

After 3 hours ship going in north east will be at a distance

Speed of ship A = 24km/hr

Speed of ship B = 32km/hr

Distance travelled by ship A after 3 hours = 24 × 3 = 72 km

Distance travelled by ship B after 3 hours = 32 × 3 = 96 km

We have to find the distance between the ships :

See the figure :

Now in Δ EFG,

EF is the distance traveled by ship A

And EG is the distance traveled by ship B

we have to find FG,

Applying cosine formula, we have-

Cos E =

Cos E = Cos 90° = 0

∴ FG² = EF² + EG²

⇒ FG =

=

∴ distance between ships after 3 hours = 120 KM ….ans

The keypoint to solve the problem:

The idea of cosine formula –

Cos C = Cos A = Cos B =

According to the question:

One ship goes in north east direction while other in southeast direction.

After 3 hours ship going in north east will be at a distance

Speed of ship A = 24km/hr

Speed of ship B = 32km/hr

Distance travelled by ship A after 3 hours = 24 × 3 = 72 km

Distance travelled by ship B after 3 hours = 32 × 3 = 96 km

We have to find the distance between the ships :

See the figure :

Now in Δ EFG,

EF is the distance traveled by ship A

And EG is the distance traveled by ship B

we have to find FG,

Applying cosine formula, we have-

Cos E =

Cos E = Cos 90° = 0

∴ FG² = EF² + EG²

⇒ FG =

=

∴ distance between ships after 3 hours = 120 KM ….ans

Given,

a=1, b=2 and ∠C=60°

By Cosine law,

2=1+4-c²

c² =5-2

c² =3

By Heron’s Law,

Area of Triangle,

sq. units

Given,

c=1, b=√3 and ∠A=30°

By Cosine law,

3=1+3-a²

a²=1

a=1

Given,

By Sine Law,

By Cosine law,

As it given

So,

For the above Equation to be true,

(c² -a² )(sin (C) )² =0

So

(c² -a² )=0

c=a

Hence Proved.

We know,

Given,

So

Now by putting cosine law

By Cosine law,

a² =400+441-672

=169

a=13

According to the product of the roots

But by sine law we know,

and

So,

Therefore,

So,

sin C=1

C=90°

Given a=8, b=10,c=12

Now by putting cosine law

By Cosine law,

=82.81924422

Therefore, C=2×A and λ=2

Lets say

a = 2x, b=√6x and c=(√3-1)x

where x is any constant

Now by putting cosine law

By Cosine law,

Putting Cosine law for B,

Putting Cosine law for C,

So, The largest Angle is 120° i.e. ∠B

According to sine law,

So,

Substituting in the given equation,

cot A= cot B=cot C

Or

tan A=tan B=tan C

Which implies

A=B=C

And we know

A+B+C=180°

So,

A=B=C=60°

Lets consider a equilateral as it is given any triangle,

So Here,

A=B=C=60°

Which implies

sin (B – C)=sin (60-60)

sin 0=0

sin (C-A)=sin(60-60)

sin 0=0

sin (A-B)=sin (60-60)

sin 0=0

Therefore,

The Equation will be,

a sin (B – C)+ b sin (C – A)+ c sin (A – B)

a sin 0+ b sin 0+ c sin 0=0

a(sinB - sinC) + b(sinC - sinA) + c(sinA - sinB)

We know,

By Sine law,

a(Kb-Kc)+b(Kc-Ka)+c(Ka-Kb)

K(ab-ac+bc-ab+ac-bc)=0

a²(sin B – sin C) + b²(sin C – sin A) + c²(sin A – sin B)

a²sinB – a²sinC + b²sinC – b²sinA + c²sinA – c²sinB

a²sinB – c²sinB + b²sinC – a²sinC + c²sinA – b²sinA

(a² – c²)sin B + (b² – a²)sin C + (c² – b²)sin A

By Sine law,

(a² – c²)bk + (b² – a²)ck + (c² – b²)ak

a²bk – c²bk + b²ck – a²ck + c²ak – b²ak

Considering it as equilateral,

a=b=c

a²bk – c²bk + b²ck – a²ck + c²ak – b²ak=0

Option D

∠A = 180-(60+75) = 45°

By Sine law,

Option B

By Cosine law,

A=30°

Putting Cosine law for B,

B=60°

Putting Cosine law for C,

C=90°

So, The largest Angle is ∠C i.e. 90° or .

Option C

By Cosine law,

and and

So,ATQ

Option C

We know by cosine law,

3c = c² -7

c² -3c-7=0

Option C

ac + bc – c² + a² + ab - ac + ab + b² - bc - ab=0

– c² + a² + ab + b² =0

a² + b² – c² = - ab

By Cosine law,

Option C

A+B+C=180°

But We Know,

So the equation will be,

a²+c²-b² = c²+a²-b²

Option B

we know by Cosine Law,

and

Option B