SOLUTION: Find to the nearest integer the area of an isosceles triangle if the measure of the vertex angle is 42 degrees and the measure of each of the congruent sides is 12.

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Question 403821: Find to the nearest integer the area of an isosceles triangle if the measure of the vertex angle is 42 degrees and the measure of each of the congruent sides is 12.
Found 2 solutions by Alan3354, richard1234:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
s = 2*12*sin(42/2)
s =~ 8.6 = base
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h = 12*sin((180-42)/2) = 12*sin(69)
h =~ 11.2
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Area = bh/2
Area =~ 48 sq units

Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
If a triangle has sides a,b and the angle between the two sides is theta, then

Area+=+a%2Ab%2Asin%28theta%29%2F2

Here, we have two congruent sides of 12, and the angle in between is 42 degrees. Thus,

Area+=+12%2A12%2Asin%2842%29%2F2+=+72+sin+%2842%29 = 48.177... = 48.