hi I reposted this question to see if I would get the answer the person who posted this one last time says that 9sqrt3 is wrong (and I check if they were right) do here is the question again
Given: ∆PQR, m∠R = 90°
m∠PQR = 75°
M ∈ PR (meaning that m in on segment PR) , MP = 18
m∠MQR = 60°
Find: RQ
Thanks (:
To find RQ, we can use the properties of a right triangle, specifically the trigonometric ratios.
Given that m∠R = 90° and m∠PQR = 75°, we can find m∠PRQ by subtracting the two angles from 180° (since the sum of the angles in a triangle is 180°).
m∠PRQ = 180° - 90° - 75°
= 15°
We can use the trigonometric ratio tangent (tan) to find RQ since we have the opposite side (MP = 18) and the adjacent angle (m∠PRQ = 15°).
tan(m∠PRQ) = opposite/adjacent
tan(15°) = RQ/18
To solve for RQ, we can multiply both sides of the equation by 18:
18 * tan(15°) = RQ
Now, let's calculate the value of RQ:
RQ ≈ 4.30
Therefore, RQ is approximately 4.30.
I made your sketch and filled in all the angles .
From triangle MPQ
QM/sin15 = 18/sin60
QM = 18sin15/(√3/2) = 16sin15/√3
in triangle RMQ
cos15 = QR/QM
QR = QM cos15
= (36sin15)/√3 * cos15
= 18(2sin15cos15)/√3
= 18(sin30)/√3
= 18(1/2) / √3
= 9/√3
= 9/√3 * √3/√3
= 9√3 /3
= 3√3