Question

Can you check my answer. Use trig to find the legs of ABC if the base is AC and the height is BC (Round to nearest tenth)

Answer 1

By definition,

`\sin (angle)=\frac{\text{ opposite side}}{\text{ hypotenuse}}`

From the picture,

`\begin{gathered} \sin (15^o)=(BC)/(60) \\ \sin (15^o)\cdot60=BC \\ 15.5\text{ ft = BC} \end{gathered}`

By definition,

`\cos (angle)=\frac{\text{ adjacent side}}{\text{ hypotenuse}}`

From the picture,

`\begin{gathered} \cos (15^o)=(AC)/(60) \\ \cos (15^o)\cdot60=AC \\ 58\text{ ft = AC} \end{gathered}`

AC is the base of the triangle and BC its height. Using the calculated values, the area of the triangle is:

`\begin{gathered} A=\text{base}\cdot\text{height}\cdot(1)/(2) \\ A=\text{58}\cdot\text{15.5}\cdot(1)/(2) \\ A=449.5^{}\text{ ft}^2 \end{gathered}`