SOLUTION: Given: m∠A = α = 42º Area of ∆ACK = 50 cm2 Find: Radius R https://lh3.googleusercontent.com/dY98UTjjE5Qn_2buhwOR56MGdOGivrHZ8H7QATyUrCR5oO5MAu4SqE-U6qBF1f7As

Algebra ->  Trigonometry-basics -> SOLUTION: Given: m∠A = α = 42º Area of ∆ACK = 50 cm2 Find: Radius R https://lh3.googleusercontent.com/dY98UTjjE5Qn_2buhwOR56MGdOGivrHZ8H7QATyUrCR5oO5MAu4SqE-U6qBF1f7As      Log On


   

Question 1150398: Given: m∠A = α = 42º
Area of ∆ACK = 50 cm2
Find: Radius R
https://lh3.googleusercontent.com/dY98UTjjE5Qn_2buhwOR56MGdOGivrHZ8H7QATyUrCR5oO5MAu4SqE-U6qBF1f7AsJxoOg=s95
Answer by Edwin McCravy(20064) About Me  (Show Source):
You can put this solution on YOUR website!
Given: m∠A = α = 42º
       Area of ∆ACK = 50 c



∆ACK is a right triangle because it is inscribed in a semicircle.

Therefore its area is half the product of its legs:

A=expr%281%2F2%29%28AC%29%28CK%29

AC%2FAK=cos%2842%5Eo%29, AC=AK%2Acos%2842%5Eo%29,

CK%2FAK=sin%2842%5Eo%29, CK=AK%2Asin%2842%5Eo%29,

AK=diameter=2R

AC=2R%2Acos%2842%5Eo%29,  CK=2R%2Asin%2842%5Eo%29,

Substituting in 

A=expr%281%2F2%29%28AC%29%28CK%29

50=expr%281%2F2%29%282R%2Acos%2842%5Eo%29%5E%22%22%29%282R%2Asin%2842%5Eo%29%5E%22%22%29

You finish.  Simplify that and solve for R.

Answer: R ≈ 7.090515777

Edwin