Chapter 8.6, Problem 21PPS

To determine

To calculate:

The distance between $A$ and $C$ :

Answer to Problem 21PPS

The value of $x$ is $2.8$ .

Explanation of Solution

Given information:

  

Concept used:

Law of Sines:

  $a=b\frac{\sin \alpha }{\sin \beta }$

Calculation:

  

The sum of the angles of the triangle is ${180}^{\circ }$ .

  $m\angle C={180}^{\circ }-\left(40+95\right)=45$

Since, the given two angles and one side set up a proportion using the Law of Sines:

Law of Sines:

  $a=b\frac{\sin \alpha }{\sin \beta }\mathrm{...}\left(1\right)$

Put the values in equation (1):

  $\frac{\sin 95}{x}=\frac{\sin 45}{2}$

Apply the property of cross products:

  $x\cdot \sin 45=2\cdot \sin 95$

Divide both the sides by $\sin 45$ :

  $\begin{array}{l}x=\frac{2\sin 95}{\sin 45}\\ x\approx 2.8\end{array}$

The value of $x$ is $2.8$ .