Chapter 9, Problem 5CUR

To determine

To find: each leg of the triangle inscribed in circle.

Answer to Problem 5CUR

The lengths of the triangle are $7,7\sqrt{3}\&14.$

Explanation of Solution

Given information:

The triangle is ${30}^0-{60}^0-{90}^0$ .

Radius of the circle $=7$

Formula used:

  $\cos \theta =\frac{\text{Adjacent side}}{\text{Hypotenuse}}$

  $\sin \theta =\frac{\text{Opposite side}}{\text{Hypotenuse}}$

Calculation:

As the given triangle is a right angled triangle, the triangle will be inscribed in a semicircle.

Consider below diagram:

  

Here, $MN$ is the diameter of the above circle.

  $MN=2(7)=14$

  $\cos \theta =\frac{\text{Adjacent side}}{\text{Hypotenuse}}$

  $\begin{array}{l}\text{cos }60°=\frac{XM}{MN}\\ \frac{1}{2}=\frac{XM}{14}\\ XM=\frac{14}{2}=7\end{array}$

  $\sin \theta =\frac{\text{Opposite side}}{\text{Hypotenuse}}$

  $\begin{array}{l}\sin 60°=\frac{XN}{MN}\\ \frac{\sqrt{3}}{2}=\frac{XN}{14}\\ XN=\frac{14(\sqrt{3})}{2}=7\sqrt{3}\end{array}$

Hence, the length of each leg of the triangle are $7,7\sqrt{3}\&14.$