Some Applications Of Trigonometry: Heights And Distances

Given,

So from the ∆ABC

sec² θ = tan²θ + 1

From the Triangle,

H= 90 × sinθ

Therefore, the height of the kite is 79.41 m.

Let us draw a diagram according to the question specification so that it gives us a better understanding of the problem.

We know that

According to the question, it is asked that we find the angle of B.

So, from the diagram, we can see that,

B = 30˚

We will first draw the schematic diagram to get a better understanding.

In the given diagram the angle of elevation is .

Actually, the angle of elevation in the question asked is of the sun with respect to the end of the shadow.

Note that if the elevation was asked of any other point like from the bottom edge of the pole, the answer would have been different.

We know that,

applying this trigonometric ratio in the problem,

= 30˚

Drawing the given problem so that we get a better understanding,

In the given diagram is the required angle as asked in the problem

We know that,

= 30˚

Therefore, the required angle the ladder makes with the wall is 30˚.

Drawing the given problem so that we get a better understanding,

In the problem, it is given that,

We know that,

using this formula in the question,

(θ is the angle of elevation as discussed in question 2)

Θ = arctan( √3)
θ = 60˚.

Therefore, the angle of elevation is 60˚.

(i) From the diagram, it is clear that the angle of elevation of A with respect to C is 30˚ where BC becomes the horizontal.

(ii) DC is a vertical line drawn from A so that we get a reference line from which we can measure the depression angle.

As AD and BC are both vertical lines therefore they are parallel.

Now ∠ACB = ∠CAD as they are alternate angles along two parallel lines

therefore, ∠ CAD = 60˚

Now, from the diagram drawn before we can easily say that

The angle of depression of C at A is 60˚

(iii)

BC = 8√3m

Now,

or,

AC = 16m.

First constructing the triangle to get the diagram,

Drawing BC a horizontal line,

Drawing a locus of A from B.

But actually nothing is mentioned about BC's length.

But we only know that ∠BAC = 60˚

And knowing that it is a right-angled triangle we can say that one angle is 90˚ and another is 180° - 90° - 60° = 30°

So, two possible cases arise,

if ∠ B = 90˚ and ∠C = 30˚, (Condition 1)

then BC > 8m

if ∠B =30˚ and ∠C= 90˚, (Condition 2)

then BC < 8m

Actual understanding of this requires knowing equation of circles and lines and equating them which results in two variables and because of the two predefined conditions which are independent we get two conditions.

Solving condition 1,

Here, AB = 8m and ∠CBA= 90˚ and ∠BAC= 60˚ and ∠ACB=30˚.

Angle of elevation of A from C will actually be equal to the ∠BCA

where BC acts as the reference line from which the elevated angle is measured.

Therefore, angle of elevation of A from C = ∠ BCA = 60˚

A horizontal line parallel to BC is drawn from A such that it acts like a reference line for calculating the angle of depression of C from A.

Angle of depression of C from A = ∠ CAD

Now, ∠CAD = ∠ACB as they are alternate angles where AD and BC acts as the parallel lines.

Therefore, ∠CAD = ∠ACB = 30˚

Therefore, the angle of depression = 60˚

Now distance of B from C i.e; BC can be calculated from trigonometric ratios.

Using,

BC = 8√3

Now,

Solving for Condition 2,

Here, angle BAC = 60˚

∠ ABC = 30˚

∠ACB = 90˚.

Also, here, BC < AB.

Now, using the above formulas and concepts for these cases we will again get the corresponding answers.

Angle of elevation of A at C = 30˚

Angle of depression of C at A= 60˚

Length of BC = 4√3

From the figure we can see that for ∠RPQ, RQ is the perpendicular and PQ is the base. Applying the formula for tangent of an angle we get,

⇒

⇒

QP = 8 m.

Drawing the triangle for better reference of the problem.

Here we are actually sure that B is the 90 degrees angle as the ends of the hypotenuse can never have 90 degrees.

Also, when the ends are produced backwards (where angle A is produced backwards at an angle of 60 degrees) they seem to form a 90 degree angle.

The remaining sides length can be figured out by simple trigonometry,

Using

⇒

or, AB = 6cm.

Now, using

⇒

or, BC=

= 6√3 cm.

Drawing the given triangle so that we get a better view of the problem.

As in the question it is given that AC is the hypotenuse, therefore it is evident that angle BAC and angle ACB cannot form as the ends of the hypotenuse never form the right angle in a right-angled triangle.

With regard to the above written point, we can say that angle ABC will form the or B will form the right angle in this right-angled triangle. Also, we can say that on producing the two ends of the hypotenuse (given one predefined angle is given) we will always get the right-angled point at the intersection of these lines (where A is produced at an angle of ) which in this case is point B.

Now using the trigonometric ratio,

Drawing the diagram so that we get a better perspective of the question,

We know that from trigonometric ratios that,

Therefore, using this ratio in this problem,

Height = 20

Therefore, the height of the pole is 20m.

We know that from trigonometric ratios that,

length of ladder = 5√2

Hence, the length of the ladder is 5√2.

Drawing the given diagram so that we get a better understanding of the problem,

We have to find the angle of elevation of the end of bamboo from the end of the shadow with the horizontal as the reference line from which the angle is measured.

We know that from trigonometric ratios that,

Therefore, using this ratio in this problem,

Θ =30°

Therefore, the angle of elevation is 30°.

We have to measure the angle of elevation of the source of light from the end of the bamboo tree with the horizontal line as the reference.

i,e; We have to find ∠ LCM

But from the diagram it is clear that, ∠LCM = ∠ACB

Now, using trigonometric ratios,

Θ = 30°

Therefore, the angle of elevation of the source of light is 30°.

Drawing a diagram for a better perspective of the problem,

We know that from trigonometric ratios that,

The angle of elevation of top of the wall will be ∠ABC.

Now Let ∠ABC = θ

Therefore,

From the figure we can see that,

AC = BC = 24 m.

Therefore,

θ = arctan(1)

θ =45°

Therefore, the angle of elevation of the top of the wall from the eyes of the observer is .

A diagram of the situation explained is drawn,

We know that from trigonometric ratios that,

Therefore, the calculated height of tower is 15√3m.

A diagram of the given situation is drawn,

We know that from trigonometric ratios that,

Therefore, the calculated height of tower is m.

A diagram of the given situation is drawn,

Using a diagram to explain why angle BCA is used as the angle of elevation.

We know that from trigonometric ratios that,

Therefore, the calculated height of tower is m.

Drawing a diagram for better understanding of the problem,

We know that from trigonometric ratios that,

height of vertical wall = 1.5√3m

Therefore, the calculated height of vertical wall is 1.5√3m.

Drawing a diagram of the given condition,

Using a diagram to explain why angle BCA is used as the angle of elevation.

In the diagram θ is the angle of elevation of point C with respect to point A where AB is the reference line from which the angle is measured.

From the theory explained above it easily understood that angle SCD is the angle of elevation

But from the diagram is also seen that angle SCD= angle ACB as both represent the same angle.

height of qutab minar = 0.89 × shadow of qutab minar

height of qutab minar= 0.89 × 81

Therefore, the height of the qutab minar is 72.09 m.

Drawing the given situation for a better understanding of the question,

Using the trigonometric ratio,

Therefore, the height of the kite is 53.33m.

(no figure was given in question 5 although the question was based from diagram, had to draw the diagram based on the answer given.)

Let in ∆ABC,

AB broken part the of Tree

AC unbroken part of the Tree

BC the distance between root and the top of the Tree

Now,

So, Height of the Tree

Therefore, the height of the tree is 51.96 m.

From the ∆ABC,

AC = 6 m.

Therefore, the length of the ladder is 6 m.

From the ∆ABC,

Therefore, the length of the rope is 10 m.

From the ∆ABC,

Therefore, distance covered by the artist to climb to the top of the pole is 20.75 m.

From the ∆ABC,

Therefore, the width of the river is 125 m.

From the ∆ABC,

Therefore, height of the plane is 92 m.

Here distance of the man from tree is given

BC = DE = 36 m

And height of man = BD = CE =1.5 m

From the ∆ABC,

Now, height of the tree = AC + CE

= 62.35 + 1.5

= 63.85

Therefore, height of the tree is 63.85 m.

Here the distance of the man from the temple is given,

BC = DE = 15 m.

And height of man =BD = CE =1 = 1.75 m

From the ∆ABC,

So,

Now, height of the temple = AC + CE = 8.66 + 1.75 = 10.41

Therefore, height of the temple is 10.41 m.

Here, point distance from base pf tower = BC = 10 m

From the ∆ABC,

Now, ∆DBC,

Therefore, the length of flagstaff = AB – DB = 10 – 2.67 = 7.32 m.

From the ∆ABC,

From the ∆DBC,

Now, the length of the flagstaff = AB – DB

=99.07 – 85.82

=13.25

Therefore, the length of the flagstaff is 13.25 m.

From the ∆ABC,

……….equation (i)

Now, from the ∆DBC,

Put value of BC from equation (i)

Therefore, the height of the tower is 10 m.

From the ∆DBC,

Now from the ∆ABC,

So, the height of the flagstaff

= AB – DB

=103.92 – 34.64

=69.28 m

Therefore, the height of the flagstaff is 69.28 m.

From the ∆DBC,

Now from the ∆ABC,

Put BC = DB from equation(i)

.46

Therefore, the height of pedestal is 2 m.

From the ∆ABC,

Now, from the ∆DBC,

Put value of BC from equation (i)

Therefore, height of the hill is 150 m.

From the ∆ADC,

Now, from the ∆ABC,

Put value of DC from equation(i) and BD = 20 that is given,

So

Therefore, height of the tower is 47.32 m.

From the ∆ADC,

Now, from the ∆ABC,

Put vale of DC from equation(i) and BD = 40 that is given.

So,

= 34.64

Now, put value of AC in equation(i)

Therefore, height of the tree is 34.64 m and width of the river is 20 m.

From the ∆DBC,

Now, from the ∆ABC,

=

Put the value of BC from the equation(i) and AD = 10 that is given.

So,

Therefore, the height of the tower is 5 m.

From the ∆DBC,

Now from the ∆ABC,

Put the value of DC from the equation(i)

So,

Therefore, height of the flagstaff is .

From the ∆DBC,

From the ∆ABC,

Put value of BC from equation(i),

Therefore, height of the tower is .

From the ∆ADC,

From the ∆ABC,

Put the value of DC from the equation(i)

Therefore, the height of the tower is m.

From the ∆ACD,

From the ∆ABC,

Therefore, the height of the church-spire is 70.80 m.

From the ∆ADC,

From the ∆DBC,

So, Distance between the two aeroplanes = AD +DB

= x + y = 1000+577.4

= 1577.4 m

Therefore, the distance between the two aeroplanes is 1577.4 m.

From the ∆ADC,

From the ∆ABC,

Therefore, the distance between the two ships is 73.2 m.

From the ∆ADC,

From the ∆ABC,

Therefore, the distance between two cars is 57.73 m

And car D is 28.87 m and car B is x+y = 86.6 m far from the tower.

From the ∆ADC,

From the ∆ABC,

Therefore, the height of the palm tree is 16.39 m.

From the ∆ADC,

From the ∆ABC,

= 17.32

put value of h in equation(i)

So,

Therefore, the height of tree is 17.32 m and width of the river is 10 m.

From the ∆ADC,

From the ∆ABC,

Therefore, the height of the tower is m.

From the ∆PBC,

From the ∆APB,

So, the height of the helicopter above the ground = x+10 = 20+10 = 30

Therefore, the height of the helicopter is 30 m.

In right Δ ABP, we have

…(i)

In the right Δ ABQ, we have

⇒ 100 – x = h

⇒ 100 = h + x

[from (i)]

Multiplying and divide by the conjugate of √3 + 1, we get

[∵ (a – b)(a + b) = (a² – b²)]

⇒ h = 50 (3 - √3)

⇒ h = 50 (3 – 1.732) [∵ √3 = 1.732]

⇒ h = 50 (1.268)

⇒ h = 63.4 m

Hence, the height of the aeroplane is 63.4m

In right Δ ABP, we have

⇒ x = √3h …(i)

In the right Δ ABQ, we have

⇒ 100 – x = h

⇒ 100 = h + x

⇒ 100 = h + √3h [from (i)]

⇒ 100 = h(√3 + 1)

Multiplying and divide by the conjugate of √3 + 1, we get

[∵ (a – b)(a + b) = (a² – b²)]

⇒ h = 50 (√3 – 1)

⇒ h = 50 (1.732 – 1)

⇒ h = 50 (0.732)

⇒ h = 36.6m

Hence, the height of the tree is 36.6m

In right Δ ABP, we have

⇒ x = 75.058 m

⇒ x = 75.06

In the right Δ ABQ, we have

⇒ y = 41.71 m

So, the distance between two men = x + y

= 75.06 + 41.71

= 116.77 m (approx.)

Let the two ships be at C and D with angles of depression 45° and 30° from point A.

The height of the light house, AB = 100m

In the right Δ ABD, we have

⇒ BD = 100√3 …(i)

In right Δ ABC, we have

⇒ BC = 100m

Hence, the distance between the two ships = BC + BD

= 100 + 100√3

= 100(1 + √3)

= 100(1 + 1.732)

= 100(2.732)

= 273.2 m (approx.)

Let AB be the idol and BC be the pillar

So, AB = 30 m

and BC = 15 m

Distance from the base of pillar = 15√3 m

Now, In right ΔACD, we have

⇒ tan x = tan 30°

⇒ x = 30°

Hence, the angle subtends from the base of the pillar is 30°

Let AB and CD are the two buildings and AE and CE are the ladder

Hence, AE and CE = 26 m (given)

In the right Δ ABE, we have

⇒ BE = 13 m

Now, In ΔCED, we have

⇒ DE = 13√2 m

So, the width of the lane = BE + DE

= 13 + 13√2

= 13 (1 + √2)

= 13 (1 + 1.414) [∵√2 = 1.414]

= 13 × 2.414

= 31.38

= 31.4 m (approx.)

Hence, the width of the lane is 31.4 m (approx.)

Let the height of the equal pillars AB = CD = h

Given the width of the road, BD = 64m

Let BE = x. Hence, DE = 64 – x

Now, In right ΔABE, we have

In the right ΔCDE, we have

⇒ 64 – x = √3h

⇒ h = 16√3 m

Hence, the height of the equal pillars AB = CD = 16√3m

Let the height of 1^st pillar CD = h and height of the 2^nd pillar = 2h

It is given that the distance between two vertical pillars is 100m

Now, In right ΔABX, we have

In the right ΔCDE, we have

⇒ 100 – x = √3h

⇒ h = 20√3 m

Hence, the height of the 1^st vertical pole, CD = 20√3 m

and the height of the 2^nd vertical pole, AB = 2 × 20√3 = 40√3 m

Let the height of the equal pillars AB = CD = h

Given the width of the road, BD = 30m

Let BE = x. Hence, DE = 30 – x

Now, In right ΔABE, we have

In the right ΔCDE, we have

⇒ 30 – x = √3h

⇒ h = 12.99m

Hence, the height of the equal pillars AB = CD = 12.99 m

The distance of a point from one pillar is

Let tree be AB and tower be CD

Given: Height of the tower = 50 m

Hence, CD = 50 m

The angle of elevation of the top of the tower from the bottom of a tree = 60°

Hence, ∠ CBD = 60°

The angle of elevation of the top of the tree from the foot of tower = 30°

Hence, ∠ ADB = 30°

Now, In right ΔCBD, we have

In the right ΔADB, we have

Hence, the height of the tree is

Let AB be the Flagstaff and CD be the vertical tower

let the height of the tower, CD = h

∵ AB = EC = 12 m

and the distance between Flagstaff and tower = 12 m

Hence, BC = AE = 12 m

Now, In Δ ABC, we have

⇒ tan x = tan 45°

⇒ x = 45° …(i)

Now, In ΔADE, we have

[from(i)]

⇒ DE = 12 m

Hence, the height of the tower, CD = DE + CE = 12 + 12 = 24m

Given: Height of the tower = 100 m

Hence, CD = 100 m = BE

Let the height of the rock = h

Hence, AB = h

In the right ΔABD, we have

⇒ BD = h

⇒ CE = h

In the right ΔAEC, we have

⇒ h = √3(h – 100)

⇒ h = √3h - √3 × 100

⇒ 100×√3 = √3h – h

⇒ 100 × √3 = h(√3 – 1)

Multiplying and divide by the conjugate of √3 – 1

⇒ h = 50(3 + 1.732)

⇒ h = 50(4.732)

⇒ h = 236.6 m

Hence, the height of the rock = 236.6 m (approx.)

Let building be AB and tower be CD

The height of building, AB = 7 m

Let the height of tower = CD

and, the distance between tower and building = AC

The angle of depression to top of the building, ∠QDB = 45°

Angle of depression to bottom of building, ∠QDA = 60°

In the right Δ BDP, we have

⇒ BP = DP …(i)

In the right Δ ADC, we have

[∵ AC = BP]

∵ CD = DP + PC

⇒ CD = DP + AB [∵ AB = PC]

⇒ CD = DP + 7

⇒ CD = 16.56 (approx.)

Given: Width of the road = 45m

In right Δ BCD, we have

⇒ CD = 15√3 m

∴ CD = BE = 15√3 m

Now, In ΔAED, we have

⇒ AE = 45√3 m

Now, the height of the building = AE + BE = 45√3 + 15√3

= √3(45 + 15)

= 60√3 m

[Hint: Let AB be the tower and CD be the building. We draw CE AB. According to the question,

CD = h, ∠BDE = α, ∠BCA = β

Let AB =y

Then, BE = BA — EA =y— h

Let CA = x. Then DE =x

From right BDE, .....(i)

Also, from right BCA, .....(ii)

Let AB be the tower and CD be the building.

We draw CE AB.

According to the question,

CD = h = BE

Let AB =y

Then, AE = AB — BE =y— h

Let CE =x. Then DB =x

In right ACE, we have

…(i)

Also, In right ABD,

…(ii)

From eq. (i) and (ii), we have

Hence, the height of the tower =

Let us suppose that DE = x and CD = y

Now, In right ΔBED, we have

⇒ x = 3600

In right ΔACE, we have

⇒ CE = 10800

⇒ CD + DE = 10800

⇒ y + x = 10800

⇒ y + 3600 = 10800

⇒ y = 10800 – 3600

⇒ y = 7200

∵ AB = CD = 7200

We know that,

Hence, the speed of the aeroplane is 864 km/hr

Let the usual speed of the plane = x km/hr

Total distance = 1500km

Actual Speed of the plane = (x + 250) km/hr

⇒ x² + 250x – 750000 = 0

⇒ x² + 1000x – 750x – 750000 = 0

⇒ x(x + 1000) – 750(x + 1000) = 0

⇒ (x – 750) (x + 1000) = 0

⇒ x + 1000 = 0 or x – 750 = 0

⇒ x = - 1000 or x = 750

⇒ x = 750 [∵ speed can’t be negative]

Hence, the usual speed of the aeroplane was 750km/hr

Let D and E be the initial and final positions of the plane respectively.

It is given that BD = 1500√3 m

In right ΔABD, we have

⇒ AB = 1500

In right ΔACE, we have

⇒ AC = 4500

Now, Distance = BC = AC – AB = 4500 – 1500 = 3000 m

DE = BC = 3000 m

i.e. the plane travels a distance of 3000m in 15 seconds

= 720km/hr

Hence, the speed of the aeroplane is 720km/hr.