Chapter 4.8, Problem 3QR

Solution

i.

Find the lengths of $b\text{ and }c.$

The value of $b\text{ is 25 and }c\text{ is 29}\text{.}$

Given: A figure of right-angle triangle.

  

Concept Used:

Trigonometric ratios: $\sin \theta =\frac{\text{Perpendicular }}{\text{Hypotenuse}},\tan \theta =\frac{\text{Perpendicular }}{\text{Base}}$

Calculation:

From the figure:

In $\Delta PQR$ :

  $PQ=15,QR=b,PR=c,\angle Q=90°,\text{ and }\angle R=31°$ .

Using trigonometric ratios:

  

  $\begin{array}{c}\tan \angle R=\frac{PQ}{QR}\\ \tan 31°=\frac{15}{b}\\ b\times \tan 31°=15\\ b=\frac{15}{\tan 31°}\\ b=\frac{15}{0.600860619}\\ b=24.9641922\\ b\approx 25\end{array}$

Now,

  $\begin{array}{c}\sin \angle R=\frac{PQ}{PR}\\ \sin 31°=\frac{15}{c}\\ c\times \sin 31°=15\\ c=\frac{15}{\sin 31°}\\ c=\frac{15}{0.515038075}\\ c=29.1240604\\ c\approx 29\end{array}$

Conclusion:

The value of $b\text{ is 25 and }c\text{ is 29}\text{.}$

ii.

Find the lengths of $a\text{and}b.$

The value of $\text{a is 66}\text{.7 and }b\text{ is 27}\text{.}$

Given: A figure of right-angle triangle.

  

Concept Used:

Trigonometric ratios: $\sin \theta =\frac{\text{Perpendicular }}{\text{Hypotenuse}},\cos \theta =\frac{\text{Base }}{\text{Hypotenuse}}$

Calculation:

From the figure:

In $\Delta PQR$ :

  $PQ=a,QR=b,PR=25,\angle Q=90°,\text{ and }\angle P=68°$ .

Using trigonometric ratios:

  $\begin{array}{c}\cos \angle P=\frac{PQ}{PR}\\ \cos 68°=\frac{25}{a}\\ a\times \cos 68°=25\\ a=\frac{25}{\cos 68°}\\ a=\frac{25}{0.374606593}\\ a=66.7366791\\ a\approx 66.7\end{array}$

  

Now,

  $\begin{array}{c}\sin \angle P=\frac{RQ}{PR}\\ \sin 68°=\frac{b}{25}\\ 25\times \sin 68°=b\\ b=\frac{25}{\sin 68°}\\ b=\frac{25}{0.927183855}\\ b=26.9633686\\ b\approx 27\end{array}$

Conclusion:

The value of $\text{a is 66}\text{.7 and }b\text{ is 27}\text{.}$

iii.

Find the lengths of $a,b,\text{and}c.$

The value of $\text{a is 40}\text{.3, }b\text{ is 23}\text{.7 and }c\text{ is 59}\text{.6}\text{.}$

Given: A figure of right-angle triangle.

  

Concept Used:

Trigonometric ratios: $\sin \theta =\frac{\text{Perpendicular }}{\text{Hypotenuse}},\tan \theta =\frac{\text{Perpendicular }}{\text{Base}}$

Calculation:

From the figure:

In $\Delta PQS$ :

  $PQ=28,PR=a,PR=a,PS=c,\angle Q=90°,\text{ and }\angle S=68°$ .

Using trigonometric ratios:

In $\Delta PQR$ :

  $\begin{array}{c}\tan \angle R=\frac{PQ}{QR}\\ \tan 44°=\frac{28}{QR}\\ QR\times \tan 44°=28\\ QR=\frac{28}{\tan 44°}\\ QR=\frac{28}{0.965688775}\\ QR=28.9948488\\ QR\approx 29\end{array}$

  

Now,

  $\begin{array}{c}\sin \angle R=\frac{PQ}{PR}\\ \sin 44°=\frac{28}{a}\\ a\times \sin 44°=28\\ a=\frac{28}{\sin 44°}\\ a=\frac{28}{0.69465837}\\ a=40.3075831\\ a\approx 40.3\end{array}$

Here, In $\Delta PQS$ :

  $\begin{array}{c}\tan \angle S=\frac{PQ}{QS}\\ \tan 28°=\frac{28}{QS}\\ QS\times \tan 28°=28\\ QS=\frac{28}{\tan 28°}\\ QS=\frac{28}{0.531709432}\\ QS=52.660341\\ QS\approx 52.7\end{array}$

So,

  $\begin{array}{c}QS=QR+RS\\ 52.7=29+b\\ b=52.7-29\\ b=23.7\end{array}$

Again in $\Delta PQS$ :

  $\begin{array}{c}\sin \angle S=\frac{PQ}{PS}\\ \sin 28°=\frac{28}{c}\\ c\times \sin 28°=28\\ c=\frac{28}{\sin 28°}\\ c=\frac{28}{0.469471563}\\ c=59.6415251\\ c\approx 59.6\end{array}$

Conclusion:

The value of $\text{a is 40}\text{.3, }b\text{ is 23}\text{.7 and }c\text{ is 59}\text{.6}\text{.}$

iv.

Find the lengths of $a,b,\text{and}c.$

The value of $a\text{ is 28}\text{.3, }b\text{ is 16 and }c\text{ is 40}\text{.8}\text{.}$

Given: A figure of right-angle triangle.

  

Concept Used:

Trigonometric ratios: $\sin \theta =\frac{\text{Perpendicular }}{\text{Hypotenuse}},\tan \theta =\frac{\text{Perpendicular }}{\text{Base}}$

Calculation:

From the figure:

In $\Delta PQS$ :

  $PQ=28,PR=a,PR=a,PS=c,\angle Q=90°,\text{ and }\angle S=68°$ .

Using trigonometric ratios:

In $\Delta PQR$ :

  $\begin{array}{c}\tan \angle R=\frac{PQ}{QR}\\ \tan 48°=\frac{21}{QR}\\ QR\times \tan 48°=21\\ QR=\frac{21}{\tan 48°}\\ QR=\frac{21}{1.11061251}\\ QR=18.9084849\\ QR\approx 18.90\end{array}$

  

Now,

  $\begin{array}{c}\sin \angle R=\frac{PQ}{PR}\\ \sin 48°=\frac{21}{a}\\ a\times \sin 48°=21\\ a=\frac{21}{\sin 48°}\\ a=\frac{21}{0.743144825}\\ a=28.2582843\\ a\approx 28.3\end{array}$

Here, In $\Delta PQS$ :

  $\begin{array}{c}\tan \angle S=\frac{PQ}{QS}\\ \tan 31°=\frac{21}{QS}\\ QS\times \tan 31°=21\\ QS=\frac{21}{\tan 31°}\\ QS=\frac{28}{0.600860619}\\ QS=34.9498691\\ QS\approx 34.9\end{array}$

So,

  $\begin{array}{c}QS=QR+RS\\ 34.9=18.9+b\\ b=34.9-18.9\\ b=16\end{array}$

Again in $\Delta PQS$ :

  $\begin{array}{c}\sin \angle S=\frac{PQ}{PS}\\ \sin 31°=\frac{21}{c}\\ c\times \sin 31°=21\\ c=\frac{21}{\sin 31°}\\ c=\frac{21}{0.515038075}\\ c=40.7736846\\ c\approx 40.8\end{array}$

Conclusion:

The value of $a\text{ is 28}\text{.3, }b\text{ is 16 and }c\text{ is 40}\text{.8}\text{.}$