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

geometry 
Steve
from what you know about 306090 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
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