Chapter 13.1, Problem 51PPS

To determine

Calculate the missing length and the missing angles of the triangle ABC.

Answer to Problem 51PPS

The angle is ${59}^{\circ }$ and the two sides are $31.6$ and $36.9$ .

Explanation of Solution

Given:

It is given in the question that the $m\angle B={31}^{\circ }\&b=19$ and $\angle C={90}^{\circ }$ .

Concept Used:

In this, use the concept of angle sum property and of trigonometric ratios and the formula will be, $\sin \theta =\frac{p}{h},\cos \theta =\frac{b}{h},\tan \theta =\frac{p}{b},\csc \theta =\frac{h}{p},\sec \theta =\frac{h}{b},\cot \theta =\frac{b}{p}$ and may also use the pythgoras theorem.

Calculation: Since, $\angle B={31}^{\circ }$ and $\angle C={90}^{\circ }$

By angle sum property,

  $\begin{array}{l}\angle A+\angle B+\angle C={180}^{\circ }\\ \angle A+{31}^{\circ }+{90}^{\circ }={180}^{\circ }\\ \angle A=180-(90+31)\\ \angle A=180-121\\ \angle A={59}^{\circ }\end{array}$

Now, using trigonometric ratios,

  $\begin{array}{l}\sin \theta =\frac{p}{h}\\ \sin {31}^{\circ }=\frac{19}{c}\\ 0.514=\frac{19}{c}\\ c=\frac{19}{0.514}\\ c\approx 36.9\end{array}$

Now, use pythagoras theorem,

  $\begin{array}{l}{a}^2+b^2=c^2\\ a^2+{19}^2={36.9}^2\\ a^2=1361.6-361\\ a^2=1000.6\\ a=\sqrt{1000.6}=31.6\end{array}$

Conclusion:

The angle is ${59}^{\circ }$ and the two sides are $31.6$ and $36.9$ .