Pls solve the 3rd question

Dear Student,

\tan 45 ° = \frac{B Q}{P Q} 1 = \frac{B Q}{P Q} H e n c e, P Q = B Q (1) \tan 60 ° = \frac{R Q}{P Q} = \frac{R B + B Q}{P Q} = \frac{20 + P Q}{P Q} (F r o m 1) \sqrt{3} P Q = 20 + P Q P Q = \frac{20}{\sqrt{3} - 1} = 27.32 B Q = 27.32

Hope this information will clear your doubts about topic.

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.

Regard

View Full Answer