Chapter 9.10, Problem 21PSD

a.

To determine

To calculate: The measure of angle R if side PQ is 10.

Answer to Problem 21PSD

The measure of angle R is ${52}^{\circ }$ .

Explanation of Solution

Given information:

In triangle PQR ,

Side PQ = 10,

Side PR = 12.

∠ Q = ${70}^{\circ }$

Formula used:

  $\sin \theta =\frac{oppositeside}{hypotenuseside}$

Calculation:

  

In ΔPQR,

  $\begin{array}{l}\sin {70}^{\circ }=\frac{x}{10}\\ \left(0.9397\right)\left(10\right)=x\\ 9.397=x\\ \sin \angle R=\frac{9.397}{12}=0.7831\\ \angle R={52}^{\circ }\end{array}$

b.

To determine

To find: The side QR if side angle Q is ${70}^{\circ }$ .

Answer to Problem 21PSD

The side QR is $11$ .

Explanation of Solution

Given information:

In triangle PQR ,

Side PQ = 10,

Side PR = 12.

∠ Q = ${70}^{\circ }$

Formula used:

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

  

In right angle triangle,

  $a^2+b^2\text{= }{c}^2$

Calculation:

  

In ΔPQR,

  $\begin{array}{l}\sin {70}^{\circ }=\frac{x}{10}\\ \left(0.9397\right)\left(10\right)=x\\ 9.397=x\end{array}$

Side PQ = 10.

Side QS can be calculated by applying Pythagoras Theorem.

In right angled triangle PSQ , we get

  $\begin{array}{l}{\left(PQ\right)}^2={\left(PS\right)}^2+{\left(SQ\right)}^2\\ {\left(10\right)}^2={\left(9.397\right)}^2+{\left(SQ\right)}^2\\ {\left(SQ\right)}^2=100-88.30\\ {\left(SQ\right)}^2=11.69\\ SQ=\sqrt{11.69}\approx 3.41\end{array}$

Side PR = 12.

Side SR can be calculated by applying Pythagoras Theorem.

In right angled triangle PSR , we get

  $\begin{array}{l}{\left(PR\right)}^2={\left(PS\right)}^2+{\left(SR\right)}^2\\ {\left(12\right)}^2={\left(9.397\right)}^2+{\left(SR\right)}^2\\ {\left(SR\right)}^2=144-88.30\\ {\left(SR\right)}^2=55.7\\ SR=\sqrt{55.7}\approx 7.5\end{array}$

  $\begin{array}{l}QR=QS+SR\\ QR=3.4+7.5\approx 11\end{array}$

c.

To determine

To show: The ratio of side PR to sine of angle Q is equal to ratio of side PQ to sine of angle R .

Explanation of Solution

Given information:

In triangle PQR ,

Side PQ = 10,

Side PR = 12.

∠ Q = ${70}^{\circ }$

Formula used:

  $\sin \theta =\frac{oppositeside}{hypotenuseside}$

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

  

In right angle triangle,

  $a^2+b^2\text{= }{c}^2$

Proof:

  

In ΔPQR,

  $\begin{array}{l}\sin {70}^{\circ }=\frac{x}{10}\\ \left(0.9397\right)\left(10\right)=x\\ 9.397=x\\ \sin \angle R=\frac{9.397}{12}=0.7831\\ \angle R={52}^{\circ }\end{array}$

Side PQ = 10.

Side QS can be calculated by applying Pythagoras Theorem.

In right angled triangle PSQ , we get

  $\begin{array}{l}{\left(PQ\right)}^2={\left(PS\right)}^2+{\left(SQ\right)}^2\\ {\left(10\right)}^2={\left(9.397\right)}^2+{\left(SQ\right)}^2\\ {\left(SQ\right)}^2=100-88.30\\ {\left(SQ\right)}^2=11.69\\ SQ=\sqrt{11.69}\approx 3.41\end{array}$

Side PR = 12.

Side SR can be calculated by applying Pythagoras Theorem.

In right angled triangle PSR , we get

  $\begin{array}{l}{\left(PR\right)}^2={\left(PS\right)}^2+{\left(SR\right)}^2\\ {\left(12\right)}^2={\left(9.397\right)}^2+{\left(SR\right)}^2\\ {\left(SR\right)}^2=144-88.30\\ {\left(SR\right)}^2=55.7\\ SR=\sqrt{55.7}\approx 7.5\end{array}$

  $\begin{array}{l}QR=QS+SR\\ QR=3.4+7.5\approx 11\end{array}$

  $\frac{PR}{\sin \angle Q}=\frac{12}{\sin {70}^{\circ }}=\frac{12}{0.9397}=12.70----Equation1$

  $\frac{PQ}{\sin \angle R}=\frac{10}{\sin {52}^{\circ }}=\frac{10}{0.7880}=12.70----Equation2$

From Equation 1 and Equation 2, we get

  $\frac{PR}{\sin \angle Q}=\frac{PQ}{\sin \angle R}$

d.

To determine

To show: The ratio of side PR to sine of angle Q is equal to ratio of side PQ to sine of angle R if triangle PQR is acute triangle.

Explanation of Solution

Given information:

In triangle PQR ,

Side PQ = 10,

Side PR = 12.

∠ Q = ${70}^{\circ }$

Formula used:

  $\sin \theta =\frac{oppositeside}{hypotenuseside}$

Proof:

  

In ΔPQS,

  $\begin{array}{l}\sin \angle Q=\frac{x}{PQ}\\ \left(PQ\right)\times \left(\sin \angle Q\right)=x----Equation1\end{array}$

In ΔPRS,

  $\begin{array}{l}\sin \angle R=\frac{x}{PR}\\ \left(PR\right)\times \left(\sin \angle R\right)=x----Equation2\end{array}$

From Equation 1 and Equation 2, we get

  $\begin{array}{l}PQ\times \sin \angle Q=PR\times \sin \angle R\\ \frac{PR}{\sin \angle Q}=\frac{PQ}{\sin \angle R}\end{array}$