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This question can be solved by exterior angle property also.

hope it will satisfy you

hope it will satisfy you

Let D be a point on ray BC such that ∠ACD is exterior angle of ΔABC at point C.

Thus, according to question, m∠ACD = b.

Now, for ΔABC,

m∠A + m∠B + m∠C = 180

∴ 70 + a + a = 180

∴ 70 + 2a = 180

∴ 2a = 180 - 70

∴ 2a = 110

∴

We know that measure of an exterior angle is the sum of measures of its interior opposite angles.

In this case ∠ACD is exterior angle and its interior opposite angles are ∠CAB and ∠ABC.

∴ m∠ACD = m∠CAB + m∠ABC

∴ b = 70 + a

∴ b = 70 + 55

∴

Thus, a = 55° and b = 125°