Theorems On Triangles

(1)

As we know, if one of the angles of the triangle is 90° then the triangle is right angle triangle.

Hence, The given triangle is a right angle triangle.

(1) – c

(2)

As we know, if one of the angles of the triangle is greater than 90° then the triangle is an obtuse angle triangle.

Hence, The given triangle is an obtuse angle triangle.

(2) – d

(3)

As we know, if all the sides of a triangle are of equal length then the triangle is an equilateral triangle.

Hence, The given triangle is a equilateral triangle.

(3) – a

(4)

As we know, if all the angles of the triangle are less than 90° then the triangle is acute angle triangle.

Hence, The given triangle is an acute angle triangle.

(4) – b

(i) Given: three sides of the triangle: 3cm, 4cm and 5cm.

As we know, if all the sides of a triangle are of different lengths then the triangle is a scalene triangle.

So, the given triangle is scalene triangle.

(ii) Given: three sides of triangle: 4cm, 7cm and 5cm.

As we know, if all the sides of a triangle are of different lengths then the triangle is a scalene triangle.

So, the given triangle is scalene triangle.

(iii) Given: three sides of triangle: 3.5cm, 4.5cm and 6cm.

As we know, if all the sides of a triangle are of different lengths then the triangle is a scalene triangle.

So, the given triangle is scalene triangle.

(iv) Given: three sides of triangle: 6.5cm, 6.5cm and 4cm.

As we know, if two of the sides of a triangle are of equal length then the triangle is an isosceles triangle.

So, the given triangle is isosceles triangle.

(v) Given: three sides of triangle: 3cm, 5.6cm and 4.3cm.

As we know, if all the sides of a triangle are of different lengths then the triangle is a scalene triangle.

So, the given triangle is scalene triangle.

(vi) Given: three sides of triangle: 2.5cm, 4.1cm and 3.2cm.

As we know, if all the sides of a triangle are of different lengths then the triangle is a scalene triangle.

So, the given triangle is scalene triangle.

(vii) Given: three sides of triangle: 5cm, 9cm and 6cm.

As we know, if all the sides of a triangle are of different lengths then the triangle is a scalene triangle.

So, the given triangle is scalene triangle.

(viii) Given: three sides of triangle: 3cm, 3cm and 3cm.

As we know, if all the sides of a triangle are of equal length then the triangle is an equilateral triangle.

So, the given triangle is equilateral triangle.

(ix) Given: three sides of triangle: 5cm, 5cm and 3.5cm.

As we know, if two of the sides of a triangle are of equal length then the triangle is an isosceles triangle.

So, the given triangle is isosceles triangle.

(x) Given: three sides of triangle: 6cm, 6cm and 8cm.

As we know, if two of the sides of a triangle are of equal length then the triangle is an isosceles triangle.

So, the given triangle is isosceles triangle.

Given: ∠A = 55° and ∠B = 40

As we know, using theorem (1), in any triangle, sum of the three interior angles is 180 °.

So,

∠A + ∠B + ∠ C = 180°

⇒ 55° + 40° + ∠ C = 180°

⇒ ∠ C = 180° - 55° - 40°

⇒ ∠ C = 85°

Given:

Let ∠A = 35° and ∠B = 90° (right angle)

As we know, using theorem (1), in any triangle, sum of the three interior angles is 180 °.

So,

∠A + ∠B + ∠ C = 180°

⇒ 35° + 90° + ∠ C = 180°

⇒ ∠ C = 180° - 35° - 90°

⇒ ∠ C = 55°

Given: vertex angle

let ∠A = 55°

in an isosceles triangle opposite angles of opposite sides are equal.

Hence,

Let ∠B = ∠ C = x

As we know, using theorem (1), in any triangle, sum of the three interior angles is 180 °.

So,

∠A + ∠B + ∠ C = 180°

⇒ 50° + x + x = 180°

⇒ 2x = 180° - 50°

⇒ 2x = 130°

⇒ x = 65°

So, ∠B = ∠ C = 65°.

Given: the ratio of angles as 1:2:3

So, let the angle,

∠A = x, ∠B = 2x, ∠ C = 3x

As we know, using theorem (1), in any triangle, sum of the three interior angles is 180 °.

So,

∠A + ∠B + ∠ C = 180°

⇒ x + 2x + 3x= 180°

⇒ 6x = 180°

⇒ x = 30°

So, the angles are,

∠A = x = 30°

∠B = 2x = 2 × 30° = 60°

∠ C = 3x = 3 × 30° = 90°

Given:

∠A = x + 15, ∠B = x - 15, ∠ C = x + 30

As we know, using theorem (1), in any triangle, sum of the three interior angles is 180°.

So,

∠A + ∠B + ∠ C = 180°

⇒ x +15 + x – 15 + x + 30 = 180°

⇒ 3x + 30° = 180°

⇒ 3x = 180° - 30°

⇒ 3x = 150°

⇒ x = 50°

So, the angles are,

∠A = x + 15 = 50° + 15° = 65°

∠B = x - 15 = 50° - 15° = 35°

∠ C = x + 30 = 50° + 30° = 80°

Given:

Let the angles are :

∠A = x, ∠B = x + 10, ∠ C = x + 20

As we know, using theorem (1), in any triangle, sum of the three interior angles is 180°.

So,

∠A + ∠B + ∠ C = 180°

⇒ x + x + 10 + x + 20 = 180°

⇒ 3x + 30° = 180°

⇒ 3x = 180° - 30°

⇒ 3x = 150°

⇒ x = 50°

So, the angles are,

∠A = x = 50°

∠B = x + 10 = 50° + 10° = 60°

∠ C = x + 20 = 50° + 20° = 70°

Given: exterior angles: ∠ ABD = 104° and ∠ ACE = 136°

As D, B and C all lie on the same line.

So,

∠ ABD + ∠ ABC = 180°

⇒ 104° + ∠ ABC = 180°

⇒ ∠ ABC = 180° - 104°

⇒ ∠ ABC = 76°

Similarly, As E, B and C all lie on the same line.

So,

∠ ACB + ∠ ACE = 180°

⇒ ∠ ACB + 136° = 180°

⇒ ∠ ACB = 180° - 136°

⇒ ∠ ACB = 44°

As we know, using theorem (1), in any triangle, sum of the three interior angles is 180°.

So,

∠ABC+ ∠ACB + ∠CAB = 180°

⇒ 76° + 44° + ∠ CAB = 180°

⇒ ∠ CAB = 180° - 76° - 44°

⇒ ∠ CAB = 60°

As we know, using theorem (2) i.e. if a side of the triangle is produced, the exterior angle so formed is equal to the sum of corresponding opposite interior angles.

So,

∠ ACD = ∠ BAC + ∠ ABC …(1)

∠ BAE = ∠ ABC + ∠ ACB …(2)

∠ CBF = ∠ BAC + ∠ BCA …(3)

Add (1), (2) and (3)

We get,

∠ ACD+∠ BAE+∠ CBF = ∠ BAC+∠ ABC+∠ ABC + ∠ BCA+ ∠ BAC + ∠ BCA

⇒∠ACD+∠BAE+∠CBF = 2(∠ BAC+∠ ABC∠ BCA)

As we know, using theorem (1), in any triangle, sum of the three interior angles is 180°.

So,

∠ABC+ ∠BAC + ∠BCA = 180°

⇒ ∠ACD+∠BAE+∠CBF = 2(180°)

⇒ ∠ACD+∠BAE+∠CBF = 360°

Hence proved.

(i) Given:

As AB = AC (isosceles triangle)

So,

∠ ABC = ∠ ACB = 50°

As B, C and D lie on the same line.

So,

∠ ACB + ∠ ACD = 180°

⇒ 50° + x = 180°

⇒ x = 180° - 50°

⇒ x = 130°

(ii)

As B, A and D lie on the same line.

So,

∠ BAC + ∠ CAD = 180°

⇒ ∠ BAC + 130° = 180°

⇒ ∠ BAC = 180° - 130°

⇒ ∠ BAC = 50°

As we know, using theorem (2) i.e. if a side of triangle is produced, the exterior angle so formed is equal to the sum of corresponding opposite interior angles.

Hence,

∠ ABE = ∠ BAC + ∠ ACB

⇒ 106° = 50° + x

⇒ x = 106° - 50°

⇒ x = 56°

(iii)

∠ BAC = ∠ EAF = 65° (vertically opposite angle)

As we know, using theorem (2) i.e. if a side of triangle is produced, the exterior angle so formed is equal to the sum of corresponding opposite interior angles.

Hence,

∠ ACD = ∠ BAC + ∠ CBA

⇒ 100° = 65° + x

⇒ x = 100° - 65°

⇒ x = 35°

(iv)

As C, A and D lie on the same line.

So,

∠ CAB + ∠ BAD = 180°

⇒ ∠ CAB + 120° = 180°

⇒ ∠ CAB = 180° - 120°

⇒ ∠ CAB = 60°

As we know, using theorem (2) i.e. if a side of triangle is produced, the exterior angle so formed is equal to the sum of corresponding opposite interior angles.

Hence,

∠ ACE = ∠ CAB + ∠ ABC

⇒ 112° = 60° + x

⇒ x = 112° - 60°

⇒ x = 52°

(v)

As AB = BC (isosceles triangle)

So,

∠ BAC = ∠ BCA = 20°

As we know, using theorem (2) i.e. if a side of the triangle is produced, the exterior angle so formed is equal to the sum of corresponding opposite interior angles.

Hence,

∠ ADB = ∠ BAC + ∠ BCA

⇒ x = 20° + 20°

⇒ x = 40°

Given: ∠TQR = 40° and ∠SPR = 30°.

∠ QTR = 90° (right angle)

As we know, using theorem (1), in any triangle, sum of the three interior angles is 180 °.

So,

∠TQR + ∠QTR + ∠ TRQ = 180°

⇒ 40° + 90° + ∠ TRQ = 180°

⇒ ∠ TRQ = 180° - 40° - 90°

⇒ ∠ TRQ = 50°

⇒ ∠ TRQ =∠ TRS = 50°

Also,

∠SPR + ∠PRS + ∠ RSP = 180°

∠SPR + ∠TRS + ∠ RSP = 180°

⇒ 30° + 50° + ∠ RSP = 180°

⇒ ∠ RSP = 180° - 30° - 50°

⇒ ∠ RSP = 100°

As R, S and Q lie on the same line.

So,

∠ RSP + ∠ PSQ = 180°

⇒ ∠ PSQ + 100° = 180°

⇒ ∠ PSQ = 80°

Given: exterior angle let ∠ ABD = 120°

One interior angle let ∠ BAC = 30°

As we know, using theorem (2) i.e. if a side of the triangle is produced, the exterior angle so formed is equal to the sum of corresponding opposite interior angles.

Hence,

∠ ABD = ∠ BAC + ∠ BCA

⇒ 120° = 30° + ∠ BCA

⇒ ∠ BCA = 120° - 30°

∠ BCA = x = 90°

As we know, using theorem (1), in any triangle, sum of the three interior angles is 180°.

So,

∠BAC + ∠BCA + ∠ ABC = 180°

⇒ 30° + 90° + y = 180°

⇒ y = 180° - 30° - 90°

⇒ y = 60°

(a) Sum of the angles of a triangle is 180° which is stated by the angle sum property of the triangle.

(b) An exterior angle of a triangle is equal to the sum of the interior opposite angles which is the exterior angle property of a triangle.

(c) An exterior angle of a triangle is always larger than either of the interior opposite angles. This is because an exterior angle of a triangle is equal to the sum of the interior opposite angles.

(d) A triangle cannot have more than one right angle. This is because the sum of the angles of a triangle is 180° if two angles will be 90° then the third angle will be 0° which is not possible.

(e) A triangle cannot have more than one obtuse angle. This is because the sum of the angles of a triangle is 180° if two angles will be more than 90° then the sum will exceed 180°.

In the given ∆ABC,

∠A = 80° and AB = AC

⇒ ∠B = ∠C {angles opposite to equal sides are equal}

Also, sum of all the angles of a triangle is 180°.

⇒ ∠A + ∠B + ∠C = 180°

⇒ 2∠B = 180° - 80°

⇒ 2∠B = 100°

⇒ ∠B = ∠C = 50°

In the given ∆ABC,

∠A = 90°

⇒ ∠B = 35°

We know that sum of all the angles of a triangle is 180°.

⇒ ∠A + ∠B + ∠C = 180°

⇒ ∠C = 180° - 90° - 35°

⇒ ∠C = 55°

Given that ∠B = ∠C = 45°

We know that the sum of the angles of a triangle is 180°.

⇒ ∠A + ∠B + ∠C = 180°

⇒ ∠A = 180° - 45° - 45°

⇒ ∠A = 90°

Hence, ∆ABC is a right triangle.

In an equilateral triangle, each angle is of 60°.

Also, an exterior angle of a triangle is equal to the sum of the interior opposite angles.

⇒ Each exterior angle = 60° + 60° = 120°

We know that an exterior angle of a triangle is equal to the sum of the interior opposite angles.

⇒ ∠A + ∠B = external ∠C

∠A + ∠C = external ∠B

∠C + ∠B = external ∠A

Adding above three, we get

2∠A + 2 ∠B + 2∠C = sum of the three external angles

⇒ 2(∠A + ∠B + ∠C) = sum of the three external angles

We know that the sum of the angles of a triangle is 180°.

⇒ 2× 180° = sum of the three external angles

⇒ Sum of the three external angles = 4 × 90°

In the given ∆ABC,

∠B = 70°

We know that the sum of all the angles of a triangle is 180°.

⇒ ∠A + ∠B + ∠C = 180°

⇒ ∠B + ∠C = 180° - 70°

⇒ ∠B + ∠C = 110°

In the given ∆ABC,

∠A = 110° and AB = AC

⇒ ∠B = ∠C {angles opposite to equal sides are equal}

Also, sum of all the angles of a triangle is 180°.

⇒ ∠A + ∠B + ∠C = 180°

⇒ 2∠B = 180° - 110°

⇒ 2∠B = 70°

⇒ ∠B = ∠C = 35°

Let the given angles of a triangle be 2x, 3x and 5x.

We know that the sum of the angles of a triangle is 180°.

⇒ ∠A + ∠B + ∠C = 180°

⇒ 2x + 3x + 5x = 180°

⇒ 10x = 180°

⇒ x = 18°

So, ∠A = 2× 18° = 36°

∠B = 3× 18° = 54°

∠C = 5× 18° = 90°

It is given that the angles of a triangle are arranged in ascending order of magnitude with the difference between consecutive angles = 15°.

So, the angles be x – 15°, x and x + 15°

We know that the sum of the angles of a triangle is 180°.

⇒ x – 15° + x + x + 15° = 180°

⇒ 3x = 180°

⇒ x = 60°

So, x – 15° = 60° - 15° = 45°

x + 15° = 60° + 15° = 75°

It is given that the sum of two angles of a triangle is equal to its third angle.

In the given ∆ABC,

Let ∠A + ∠B = ∠C …(1)

We know that the sum of the angles of a triangle is 180°.

⇒ ∠A + ∠B + ∠C = 180°

⇒ ∠C + ∠C = 180° {From (1)}

⇒ 2∠C = 180°

⇒ ∠C = 90°

In a triangle ABC, it is given that

2∠A = 3∠B = 6∠C

⇒ 2∠A = 6∠C

⇒ ∠A = 3∠C

Similarly,3∠B = 6∠C

⇒ ∠B = 2∠C

We know that the sum of the angles of a triangle is 180°.

⇒ ∠A + ∠B + ∠C = 180°

⇒ 3∠C + 2∠C + ∠C = 180°

⇒ 6∠C = 180°

⇒ ∠C = 30°

So, ∠A = 3× ∠C = 90°

∠B = 2× ∠C = 60°

It is given that the angles of a triangle are x – 40°, x – 20° and x + 15°.

So, the angles be x – 15°, x and x + 15°

We know that the sum of the angles of a triangle is 180°.

⇒ x – 40° + x – 20° + x + 15° = 180°

⇒ 3x – 45° = 180°

⇒ 3x = 180° + 45°

⇒ x = 75°

In a triangle ABC, it is given that

∠A – ∠B = 15° and ∠B – ∠C = 30°

⇒ ∠A = ∠B + 15° and ∠C = ∠B – 30°

We know that the sum of the angles of a triangle is 180°.

⇒ ∠A + ∠B + ∠C = 180°

⇒ ∠B + 15° + ∠B + ∠B – 30° = 180°

⇒ 3∠B = 180° + 15°

⇒ 3∠B = 195°

⇒ ∠B = 65°

So, ∠A = ∠B + 15° =65°+15° = 80°

∠C = ∠B – 30°= 65°- 30° = 35°

In a triangle ABC, it is given that

∠A – ∠B = 20° … (1)

and ∠A + ∠B = 80° .. (2)

Add (1) and (2),

⇒ 2∠A = 100°

⇒ ∠A = 50°

So, ∠B = 80° - ∠A = 80° - 50° = 30°

We know that the sum of the angles of a triangle is 180°.

⇒ ∠A + ∠B + ∠C = 180°

⇒ 50° + 30° + ∠C = 180°

⇒ ∠C = 180° - 80°

⇒ ∠C = 100°

It is given that in a triangle ABC, ∠B = 60°, and ∠C = 80°. IB and IC are the bisectors of ∠B and ∠C respectively.

⇒ ∠IBC = 30° and ∠ICB = 40°

We know that the sum of the angles of a triangle is 180°.

So, in ∆ IBC

⇒ ∠BIC + ∠IBC + ∠ICB = 180°

⇒ ∠BIC = 180° - 40° - 30°

⇒ ∠BIC = 110°

In a triangle ABC, let ∠C be the larger angle and ∠A and ∠B be the smaller ones. It is given that

⇒ 2∠A = 2∠B = ∠C

⇒ ∠A = ∠B

And ∠C = 2∠A

We know that the sum of the angles of a triangle is 180°.

⇒ ∠A + ∠B + ∠C = 180°

⇒ ∠A + ∠A + 2∠A = 180°

⇒ 4∠A = 180 °

⇒ ∠A = 45°

So, ∠B = ∠A = 45°

∠C = 2× ∠A = 90°

In a triangle ABC, let ∠C be the third angle and ∠A and ∠B be the bigger ones. It is given that

∠A = ∠B = 2∠C

We know that the sum of the angles of a triangle is 180°.

⇒ 2∠C + 2∠C + ∠C = 180°

⇒ 5∠C = 180 °

⇒ ∠C = 36°

So, ∠B = ∠A = 2× ∠C = 72°

It is given that in ∆ ABC,

∠B = 50° and ∠A = 60°

Here, ∠ACD is an external angle. We know that an exterior angle of a triangle is always larger than either of the interior opposite angles.

⇒ ∠ACD = ∠A + ∠B = 50° + 60° = 110°

In a triangle ABC, let ∠A be the vertex angle and ∠C and ∠B be the base angles. It is given that

∠B = ∠C {∵ it is an isosceles triangle}

And ∠A = 2(∠B + ∠C)

⇒ ∠A = 2× 2 ∠B

∠A = 4 ∠B

We know that the sum of the angles of a triangle is 180°.

⇒ 4∠B + ∠B + ∠B = 180°

⇒ 6∠B = 180 °

⇒ ∠B = 30°

So, ∠C = ∠B = 30°

And ∠A = 4× ∠B = 120°

The given star is of the pentagram shape. Each angle at vertices is of 36°. So, the sum of all the angles at the five vertices of the adjoining star = 5× 36° = 180°