SOLUTION: Given: ∆AKM, R = 2, m∠A = 33°, O∈AM. Find: perimeter of ∆AKM https://lh3.googleusercontent.com/XxKrlU8bFG1SlkyIN7TnUgujgwFyLSlOVvHPOX7Z5ROrqCGAlXuvwvfMN

Algebra ->  Trigonometry-basics -> SOLUTION: Given: ∆AKM, R = 2, m∠A = 33°, O∈AM. Find: perimeter of ∆AKM https://lh3.googleusercontent.com/XxKrlU8bFG1SlkyIN7TnUgujgwFyLSlOVvHPOX7Z5ROrqCGAlXuvwvfMN      Log On


   

Question 1150085: Given: ∆AKM, R = 2,
m∠A = 33°, O∈AM.
Find: perimeter of ∆AKM
https://lh3.googleusercontent.com/XxKrlU8bFG1SlkyIN7TnUgujgwFyLSlOVvHPOX7Z5ROrqCGAlXuvwvfMN9C3S4fiiK7Zbg=s97
Answer by Edwin McCravy(20066) About Me  (Show Source):
You can put this solution on YOUR website!
 

∆AKM is a right triangle for it is inscribed in semicircle O.

AM = diameter of semi-circle O = 2R = 2(2) = 4

%22AK%22%2F%22AM%22=cos%2833%5Eo%29
AK+=+AM%2Acos%2833%5Eo%29
AK+=+4cos%2833%5Eo%29

%22KM%22%2F%22AM%22=sin%2833%5Eo%29
KM+=+AM%2Asin%2833%5Eo%29
KM+=+4sin%2833%5Eo%29

The perimeter of a triangle is the sum of the three sides.

perimeter=AK%2BKM%2BAM

perimeter=4cos%2833%5Eo%29%2B4sin%2833%5Eo%29%2B4

Approximately 9.533238412

Edwin