Question

In ∆ABC the angle bisectors drawn from vertices A and B intersect at point D. Find ∠ADB if:

Answer 1

Answer:

∠ADB = 90°+0.5γ

Step-by-step explanation:

Given that;

∠ADB =180°-(1/2 ∠A + 1/2 ∠B)----sum of interior angles of the triangle ABD add upto 180°

From the other side;

∠A + ∠B + ∠C =180° -------------(i)

Equation (i) is equal to 1/2 ∠A + 1/2 ∠B = 90°-1/2∠C thus

∠ADB = 180°-(1/2 ∠A + 1/2 ∠B) = 180°-(90°-1/2∠C) = 90°+1/2∠C

So in terms of γ where ∠C=γ

∠ADB = 90° + 1/2 γ

∠ADB = 90° + 0.5γ