Chapter 6.1, Problem 61E

To determine

Whether the given information holds true to apply law of Sines.

Answer to Problem 61E

Yes, the values are $\angle A=40°,b=11.92\text{and}c=15.56$ .

Yes, using the definitions of cosine and tangent.

Explanation of Solution

Given information:

A triangle with

  $B=50°,C=90°,\text{and }a=10.$

Since the triangle is a right triangle, finding $\angle A$ can be done using the angle sum property. Then using the law of Sines, we can find the sides $b\text{and}c$ .

That is,

  $\begin{array}{l}\angle A=180-\angle B-\angle C\\ \Rightarrow \angle A=180-50-90\\ \Rightarrow \angle A=40°\end{array}$

Then using law of Sines,

  $\begin{array}{l}\frac{\sin (A)}{a}=\frac{\sin (B)}{b}=\frac{\sin (C)}{c}\\ \Rightarrow \frac{\sin (40)}{10}=\frac{\sin (50)}{b}=\frac{\sin (90)}{c}\end{array}$

To find $b$ :

  $\begin{array}{l}\frac{\sin (40)}{10}=\frac{\sin (50)}{b}\\ \Rightarrow b=\frac{10\sin (50)}{\sin (40)}=11.92\end{array}$

To find $c$ :

  $\begin{array}{l}\frac{\sin (40)}{10}=\frac{\sin (90)}{c}\\ \Rightarrow c=\frac{10\sin (90)}{\sin (40)}=15.56\end{array}$

Yes, there is an alternate way of solving the triangle. That is using the definitions of cosine and tangent.

That is,

  $\begin{array}{l}\cos (B)=\frac{a}{c}\\ \Rightarrow \cos (50)=\frac{10}{c}\\ \Rightarrow c=\frac{10}{\cos (50)}=15.56\end{array}$

And

  $\begin{array}{l}\tan (A)=\frac{a}{b}\\ \Rightarrow \tan (40)=\frac{10}{b}\\ \Rightarrow b=\frac{10}{\tan (40)}=11.92\end{array}$