Question

In a triangle ABC, ∠C = 3∠B = 2 (∠A+ ∠B).

What are the three angles?

Answer 1

Given :-

• ∠C = 3∠B = 2(∠A+ ∠B)

To Find :-

Value of ∠A, ∠B and ∠C

Solution :-

∠C = 2(∠A + ∠B)

∠C = 2∠A + 2∠B ---- 1

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A + B + C = 180

A + B + 2A + 2B = 180 (From 1)

3A + 3B = 180 (Dividing by 3)

A + B = 60 ----2

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A + B + C = 180

60 + C = 180 (From 2)

C = 180 - 60

∠C = 120°

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2(A + B) = ∠C  (Given)

2(A + B) = 3B

2A + 2B = 3B

B = 2A ---3

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A + B = 60 (From 2)

A + 2A = 60

3A = 60

A = 60/3

∠A = 20°

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B = 2A (From 3)

B = 2(20)

∠B = 40°

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→

• ∠A = 20°

• ∠B = 40°

• ∠C = 120°

Answer 2

SOLUTION:-

Given:

$\angle C = 3 \angle B= 2( \angle A+ \angle B)$

If A, B & C are the angles of a triangle,

$\angle A + \angle B + \angle C = 180 \degree...........(1)$

Given relation between the angles is:

$\angle C = 3 \angle B= 2( \angle A+ \angle B)$

Therefore,

$3 \angle B = 2( \angle A + \angle B) \\ \\ 2 \angle A= 3 \angle B - 2 \angle B \\ \\ 2 \angle A = \angle B \\ \\ \angle A = \frac{ \angle B}{2}$

Substitute in equation (1):

$\frac{ \angle B}{2} + \angle B + 3 \angle B= 180 \degree \\ \\ \frac{ \angle B}{2} + 4 \angle B= 180 \degree \\ \\ \angle B ( \frac{1 + 8}{2}) = 180 \degree \\ \\ \angle B \times \frac{9}{2} = 180 \degree \\ \\ \angle B = \frac{180 \degree \times 2}{9} \\ \\ \angle B= 20 \degree \times 2 \\ \\ \angle B = 40 \degree$

Thus,

$\angle C =3 \angle B \\ \\ = > 3 \times 40 \degree \\ \\ = > 120 \degree \\ \\ \\ \angle A = \frac{ \angle B}{2} \\ \\ = > \frac{40 \degree}{2} \\ \\ = > \angle A = 20 \degree$

Hence,

⚫angle A = 20°

⚫Angle B = 40°

⚫Angle C= 120°

Hope it helps ☺️