Chapter 70, Problem 36A

To determine

The angle $X$ of the triangle.

The angle $Y$ of the triangle.

Answer to Problem 36A

The angle $X$ of the triangle is $=79.74°$.

The angle $Y$ of the triangle is $=71.44°$.

Explanation of Solution

Given:

A figure as above, in which different dimensions are shown.

  $\text{AB}=7.315"$, $\text{BC}=\text{7}\text{.590"}$, $\angle A=\angle X$, $\angle C=\angle Y$ and $\angle B=28.4°$.

Concept Used:

The law of cosines.

  $\begin{array}{l}{a}^2=b^2+c^2-2bc\cos A\\ b^2=a^2+c^2-2ac\cos B\\ c^2=a^2+b^2-2ab\cos C\end{array}$

or

  $\begin{array}{l}\cos A=\frac{b^2+c^2-a^2}{2bc}\\ \cos B=\frac{a^2+c^2-b^2}{2ac}\\ \cos C=\frac{a^2+b^2-c^2}{2ab}\end{array}$

The law of sines

  $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

Calculation:

As per the given problem in the $\text{ΔABC}$

  $\text{c}=7.315"$, $\text{a}=\text{7}\text{.590"}$, $\text{AC}=b$

  $\angle A=\angle X$, $\angle B=28.4°$ and $\angle C=\angle Y$.

Substitute the values in the formula $b^2=a^2+c^2-2ac\cos B$, we get

  $\begin{array}{l}$ b^2={(7.59)}^2+{(7.315)}^2-2(7.59)(7.315)\cos 28.4°$b^2=57.608+53.509-97.68\\ b^2=13.44\\ b=3.6\text{7 inch}\end{array}$

Now, substitute the values in the formula $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$, we get.

  $\frac{7.590}{\sin \angle X}=\frac{3.67}{\sin 28.4°}=\frac{7.315}{\sin \angle Y}$

By solving $\frac{7.590}{\sin \angle X}=\frac{3.67}{\sin 28.4°}$, we get.

  $\begin{array}{c}\sin \angle X=7.590\times \frac{\sin 28.4°}{3.67}\\ \angle X={\sin }^{-1}(0.984)\\ \angle X=79.74°\end{array}$

The angle $X$ of the triangle is $79.74°$.

By solving $\frac{3.67}{\sin 28.4°}=\frac{7.315}{\sin \angle Y}$, we get.

  $\begin{array}{c}\sin \angle Y=\frac{7.315}{3.67}\times \sin 28.4°\\ \angle Y={\sin }^{-1}(0.948)\\ \angle Y=71.44°\end{array}$

The angle $Y$ of the triangle is $71.44°$.

Conclusion:

Therefore, the angle $X$ of the triangle is $=79.74°$.

Therefore, the angle $Y$ of the triangle is $=71.44°$.