Question 90

In the figure if ∠ACB = 40°, ∠DPB = 120°, then find y.

Solution

$$\angle$$CPA = $$\angle$$DPB = 120° (Vertically opposite angles)

In $$\triangle$$ APC

=> $$\angle$$ ACP + $$\angle$$ CPA + $$\angle$$ PAC = $$180^{\circ}$$

=> $$\angle$$ PAC = $$180^{\circ} - 40^{\circ} - 120^{\circ} = 20^{\circ}$$

Also, $$\angle$$ PAC = $$\angle$$ PBD (Angles in the same segment)

=> $$\angle$$ PBD = $$y = 20^{\circ}$$

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