Question

In Fig. 16.139, if ∠ACB=40°, ∠DPB=120°, find ∠CBD.

Answer 1

∠CBD = 20°

Step-by-step explanation:

• Given data

        ∠ACB = 40° =∠ACP

        ∠DPB = 120° = ∠APC     (vertically opposite angle)

• Now in triangle ΔAPC, from triangle angle property

       ∠ACP +∠APC +∠CAP = 180°

       40° +120° +∠CAP = 180°

       So

       ∠CAP = 20°

• We know that $\mathbf{\Delta APC\sim \Delta BPD}$ in given figure

        So corresponding angle is also equal in similar triangle

        ∠PBD =∠CAP

        ∠PBD = 20° =∠CBD

Answer 2

Answer:

joint the point A and B

∠ACB = ∠ADB = 40°

△BDP ∠DPB + ∠PBD + ∠BDP = 180°

⇒ 120° + ∠PBD + 40° = 180°

⇒ ∠PBD = 20°

Hence,∠CBD = 20°