Question

triangle abc angle c=5 angle b=3(a+b) angle a ha to tino kono ka man kiya hoga and

Answer 1

Question :

In a traingle ABC,

∠C = 5 ∠B = 3 (∠A + ∠B).

Find ∠A, ∠B and ∠C.

Solution :

We know that the sum of the three angles of a triangle = 180°

i.e., for traingle ABC,

∠A + ∠B + ∠C = 180° ...(1)

Given that,

∠C = 5 ∠B = 3 (∠A + ∠B) ...(2)

We have

3 (∠A + ∠B) = ∠C [ by (2) ]

or, 3 (180° - ∠C) = ∠C [ by (1) ]

or, 540° - 3 ∠C = ∠C

or, 4 ∠C = 540°

∴ ∠C = 135°

Again, we have

∠C = 5 ∠B

or, 5 ∠B = 135°

∴ ∠B = 27°

From (1), we get

∠A + 27° + 135° = 180°

or, ∠A = 180° - 27° - 135°

∴ ∠A = 18°

Thus, solved.

Answer 2

solution:-

Given, angle c = 5angle b = 3(angleA+angleB)

let the angle b be x

so the angle C = 5x

5 angleB=3 (angleA + angleB)

5x=3(y+x) (since angle A=Y)

2x/3 = Y

we know that angle sum property of triangle

A+b+c (angle) X + 5 x