# geometry

Given: ∆PQR, m∠R = 90°
m∠PQR = 75°
M ∈
PR
, MP = 18
m∠MQR = 60°
Find: RQ

1. from what you know about 30-60-90 right triangles, if we let x = RQ then
MR = x√3

Now, we have

(18+x√3)/x = tan 75°

So solve for x, which is RQ.

It might help if you recall that tan75° = 2+√3

posted by Steve

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