Question

(1 point) Let @ (in radians) be an acute angle in a right triangle and let x and y, respectively, be the lengths of the sides

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Answer #1

We can draw triangle

8. ២/Y 3 X

we are given

x=8,y=5

\frac{dx}{dt}=4,\frac{dy}{dt}=-\frac{1}{6}

we can use sine formula

sin(\theta)=\frac{y}{\sqrt{89}}

we can find derivative with respect to t

\frac{d\theta}{dt}cos(\theta)=\frac{\frac{dy}{dt}}{\sqrt{89}}

now, we can plug cos value

\frac{d\theta}{dt}(\frac{8}{\sqrt{89}})=\frac{\frac{dy}{dt}}{\sqrt{89}}

\frac{d\theta}{dt}(\frac{8}{\sqrt{89}})=\frac{-\frac{1}{6}}{\sqrt{89}}

\frac{d\theta}{dt}=\frac{\frac{-\frac{1}{6}}{\sqrt{89}}}{(\frac{8}{\sqrt{89}})}

so, we get

\frac{d\theta}{dt}=-\frac{1}{48} rad/sec...........Answer

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