Question

The adjoining figure, shows two straight lines AB and CD intersecting at P. If $$\angle BPC = 4x - 5^\circ$$ and $$\angle APD = 3x + 15^\circ,$$ find $$\angle BPD$$.

Answer 1

By referring given figure,

The value of $$x$$ is calculated as,

$$3x + 15^\circ = 4x - 5^\circ$$ (By vertically opposite angles)

$$3x - 4x = -5^\circ - 15^\circ$$

$$-x = -20^\circ$$

$$x = 20^\circ$$

The value of $$\angle BPD$$ is calculated as,

The sum of adjacent angles is $$180^∘$$

$$\angle BPD+ \angle BPC = 180^\circ $$

$$\angle BPD = 180^\circ - \angle BPC$$

$$= 180^\circ - (4x - 5^\circ)$$

$$= 180^\circ - (4 \times 20^\circ - 5^\circ )$$

$$= 180^\circ - 80^\circ + 5^\circ$$

$$= 105^\circ$$