Question

Mark the correct alternative of the following.
In figure, if AB is parallel to CD, then the value of $$\angle BPE$$ is?

Answer 1

Correct option is C. $$74^o$$
In figure, AB || CD and EF intersecting AB and CD

$$\angle BPQ= \angle CQP$$ (alternate angles)

So, $$\angle BPQ = \angle CQP$$

$$5x - 14 = 3x + 34$$

$$5x - 3x = 14 + 34$$

$$2x = 48$$

$$x = 24$$

$$If \ \ x = 24$$

So, $$\angle BPF = 3x + 34 $$

$$= 3 \times 24+ 34 $$

= $$72 + 34 = 106^o$$

$$\angle BPF + \angle BPE = 180^o$$ (linear pair)

$$106^o + \angle BPE = 180^o$$

$$ \angle BPE = 74^o$$