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In a right triangle, cos (2x) = sin (8x + 5)'. Find the smaller of the triangle's two
acute angles.

In A Right Triangle Cos 2x Sin 8x 5 Find The Smaller Of The Triangles Twoacute Angles class=

Sagot :

According to the given problem,

[tex]\cos (2x)^{\circ}=\sin (8x+5)^{\circ}[/tex]

Consider the formula,

[tex]\sin (90-\theta)=\cos \theta[/tex]

Apply the formula,

[tex]\sin (90-2x)=\sin (8x+5)[/tex]

Comparing both sides,

[tex]\begin{gathered} 90-2x=8x+5 \\ 8x+2x=90-5 \\ 10x=85 \\ x=\frac{85}{10} \\ x=8.5 \end{gathered}[/tex]

Obtain the value of the two angles,

[tex]\begin{gathered} 2x=2(8.5)=17 \\ 8x+5=8(8.5)+5=73 \end{gathered}[/tex]

It is evident that the smaller angle is 17 degrees, and the larger angle is 73 degrees.

Thus, the required value of the smaller acute angle of the triangle is 17 degrees.

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