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In a right triangle, sin (8x + 6)° = cos (9x - 1)°. Find the larger of the triangle's two acute angles.​

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Answer:

the two acute angles is 46°.

Explanation:

In a right triangle, the sine of an angle is equal to the cosine of its complementary angle. So, if sin (8x + 6)° = cos (9x - 1)°, then (8x + 6)° and (9x - 1)° are complementary angles. This means that their sum is 90°. So, we can write the equation:

(8x + 6)° + (9x - 1)° = 90°

Solving for x, we get:

8x + 6 + 9x - 1 = 90 17x + 5 = 90 17x = 85 x = 5

Substituting x = 5 into either of the original expressions for the angles, we find that one acute angle is (8 * 5 + 6)° = 46° and the other is (9 * 5 - 1)° = 44°

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