Chapter 15, Problem 29CR

To determine

To find: the acute angle formed by the diagonals of the quadrilateral.

Answer to Problem 29CR

  ${60}^{\circ },{72}^{\circ },{84}^{\circ }$ .

Explanation of Solution

Given:

The quadrilateral $PQRS$ is inscribed in the circle center O.

The measures of $PQ,QR,RS,SP$ are in the ratio of $7:12:6:5$ .

Calculation:

According to the question, the quadrilateral $PQRS$ is inscribed in the circle with center O.

  

Now, let the angles in the ratio be x . then,

  $\begin{array}{l}1\text{st}\text{angle}=7x\\ 2\text{nd}\text{angle}=12x\\ 3\text{rd}\text{angle}=6x\\ \text{4th}\text{angle}=5x\end{array}$

As the sum of all triangles of the quadrilateral is ${360}^{\circ }$ . Therefore,

  $\begin{array}{l}7x+12x+6x+5x={360}^{\circ }\\ \Rightarrow 30x={360}^{\circ }\\ \therefore x=\frac{{360}^{\circ }}{30}\\ \Rightarrow x={12}^{\circ }\end{array}$

Thus, the measurement of each angles are:

  $\begin{array}{l}1\text{st}\text{angle}=7x=7\times 12={84}^{\circ }\\ 2\text{nd}\text{angle}=12x=12\times 12={144}^{\circ }\\ 3\text{rd}\text{angle}=6x=6\times 12={72}^{\circ }\\ \text{4th}\text{angle}=5x=5\times 12={60}^{\circ }\end{array}$

Therefore, the acute angle formed by the diagonals of the quadrilateral is ${60}^{\circ },{72}^{\circ },{84}^{\circ }$ .