Chapter 8.6, Problem 55HP

To determine

To calculate:

The value of $x$ in the figure at the right:

  

Answer to Problem 55HP

The value of $x$ in the figure at the right is $5.6$ .

Explanation of Solution

Given information:

  

Calculation:

The vertices are:

  

If $\Delta ABC$ has length $a$ , $b$ and $c$ representing the length of sides opposites the angles with measure $A$ , $B$ and $C$ .

Then, use the law of cosine:

  $a^2+b^2+c^2-2bc\cos A$

In right angle $\Delta ABC$ , the law of cosine:

  $\cos B=\frac{{\left(BA\right)}^2+{\left(BC\right)}^2-{\left(AC\right)}^2}{2\left(BA\right)\left(BC\right)}$

Substitute the values in law of cosine:

  $\cos B=\frac{{5.9}^2+6^2-6^2}{2\left(6\right)\left(5.9\right)}$

Then simplify:

  $\cos B=0.491667$

Definition of Inverse cosine:

  $\begin{array}{l}m\angle B={\cos }^{-1}\left(0.491667\right)\\ m\angle B=60.55\end{array}$

If $\Delta ABC$ has length $a$ , $b$ and $c$ representing the length of sides opposites the angles with measure $A$ , $B$ and $C$ .

Then, use the law of sine:

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

In right angle $\Delta ABC$ , the law of sine:

  $\frac{\sin {58}^{\circ }}{6}=\frac{\sin {60.55}^{\circ }}{x}$

The property of cross products:

  $x\sin {58}^{\circ }=6\sin {60.55}^{\circ }$

Then property of multiplication:

  $\begin{array}{l}x=\frac{6\sin {60.55}^{\circ }}{\sin {68}^{\circ }}\\ x\approx 5.6\end{array}$