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In a right triangle, sin (x+4)^\circ(x+4)



= cos (3x+2)^\circ(3x+2)


. Find the larger of the triangle’s two acute angles


Sagot :

The larger of the triangle’s two acute angles is 65 degrees

Angles in a triangle

Given the expression

sin(x+4) = cos(3x + 2)

Since sin theta = cos(90 - theta), hence;

cos(90 - (x+4)) = cos(3x +2)

cos(86 - x) = cos (3x+2)

Cancel out the cosine functions to have:

86 - x = 3x + 2

-4x = -84

x = 21

Substitute x = 21 into the expression

sin(x+4) = Sin25 =

Hence the larger of the triangle’s two acute angles is 65 degrees

Learn more on acute angles here; https://brainly.com/question/6979153

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