Question

Note: Enter your answer and show all the steps that you use to solve this problem in the space provided

Answer 1

From the given picture we can see

ACB is a right triangle at C

AC = b

CB = a

AB = c

Since mSince a = 5 ft

Then to find b and c we will use the trigonometry ratios

`\begin{gathered} \sin A=(a)/(c) \\ \sin 60=(5)/(c) \end{gathered}`

Substitute the value of sin 60

`\begin{gathered} \sin 60=\frac{\sqrt[]{3}}{2} \\ \frac{\sqrt[]{3}}{2}=(5)/(c) \end{gathered}`

By using the cross multiplication

`\begin{gathered} \sqrt[]{3}* c=2*5 \\ \sqrt[]{3}c=10 \end{gathered}`

Divide both sides by root 3

`\begin{gathered} \frac{\sqrt[]{3}c}{\sqrt[]{3}}=\frac{10}{\sqrt[]{3}} \\ c=\frac{10}{\sqrt[]{3}} \end{gathered}`

To find b we will use the tan ratio

`\begin{gathered} \tan 60=(a)/(b) \\ \tan 60=(5)/(b) \end{gathered}`

Substitute the value of tan 60

`\begin{gathered} \tan 60=\sqrt[]{3} \\ \sqrt[]{3}=(5)/(b) \end{gathered}`

Switch b and root 3

`b=\frac{5}{\sqrt[]{3}}`

The exact values of b and c are

`\begin{gathered} b=\frac{5}{\sqrt[]{3}} \\ c=\frac{10}{\sqrt[]{3}} \end{gathered}`