Solve the equation on the interval [0,2pi) 4sin^2x-3=0
1 answer:
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9514 1404 393
Answer:
(a) 42.3°
Step-by-step explanation:
Side 'a' is the shortest of three unequal sides, so angle A will be the smallest angle in the triangle. Its measure can be found from the Law of Cosines.
a² = b² +c² -2bc·cos(A)
cos(A) = (b² +c² -a²)/(2bc) = (21² +25² -17²)/(2·21·25) = 777/1050
A = arccos(777/1050) ≈ 42.3°
The measure of angle A is about 42.3°.
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<u><em>Additional comment</em></u>
The smallest angle in a triangle can never be greater than 60°. This lets you eliminate choices that exceed that value.
