Chapter 6, Problem 17T

To determine

To find: The value of angle labeled $\theta$.

Answer to Problem 17T

The value of $\theta$ is $78.6°$.

Explanation of Solution

Given:

The value of one side is $9$, second side is $6$ and third side is $8$.

Formula used:

Law of Cosines $a^2=b^2+c^2-2bc\cos A$.

Calculation:

The given triangle has three sides.

Then for SSS, use law of cosine to find the angle opposite to the first side

$a^2=b^2+c^2-2bc\cos A$

Substitute $9$ for $a$, $6$ for $b$ and $8$ for $c$.

$\begin{array}{c}{\left(9\right)}^2={\left(6\right)}^2+{\left(8\right)}^2-2\times 6\times 8\times \cos A\\ 96\cos A=100-81\\ \cos A=\frac{19}{96}\\ A={\cos }^{-1}\frac{19}{96}\end{array}$

Use the $\mathrm{ARC}$ $\mathrm{COS}$ keys on the scientific calculator with the degree mode to find the approximate value of ${\cos }^{-1}\frac{19}{96}$.

Adjust the scientific calculator for inverse functions key.

Then adjust for the degree mode.

Enter the value $\arccos \left(\frac{19}{96}\right)$.

The calculator shows value

$\begin{array}{c}\arccos \left(\frac{19}{96}\right)=78.58\\ \approx 78.6\end{array}$

The value of angle labeled $\theta$ for ${\cos }^{-1}\frac{19}{96}$ is $78.6°$.