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3 years ago

Given: ∆AKM, R = 2, m∠A = 33°, O∈ AM . Find: perimeter of ∆AKM

Mathematics

2 answers:

Gwar [14]3 years ago

8 0

Answer:

∆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

Step-by-step explanation:

sergeinik [125]3 years ago

7 0

Answer:

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Step-by-step explanation:

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