Question

In ∆ABC, ∠Α = 50°, ∠B = 70° and bisector of ∠C meets AB in D (Fig.). Measure of ∠ADC is.

(a) 50° (b) 100° (c) 30° (d) 70°

Answer 1

(b) 100^o

From the figure,

Consider the ΔABC,

We know that, sum of angles of triangle is equal to 180^o.

So, ∠Α + ∠B + ∠C = 180^o

50^o + 70^o + ∠C = 180^o

∠C + 120^o = 180^o

∠C = 180^o – 120^o

∠C = 60^o

Since CD bisects ∠C,

So, ∠DCB = ∠ΑCD = ½ ∠C

= 60^o/2

= 30^o

Now, consider ΔBDC

From exterior angle property, ∠ACD = ∠DBC + ∠DCB

∠ACD = 70^o + 30^o

= 100^o