Chapter 8.4, Problem 12WE

Solution

To complete: the given table.

d	e	f
$2\sqrt{3}$	$6$	$4\sqrt{3}$

Given:

The figure:

And the table:

d	e	F
?	$6$	?

Concept Used:

Theorem 3-11: The addition of measures of all angles of a triangle is 180.

Theorem 8-7: In a $30°-60°-90°$ triangle, the longer Side is $\sqrt{3}$ times as long as the shorter Side and the hypotenuse of right triangle are two times as long as the shorter side of triangle.

Explanation:

Label the given figure as shown below:

Apply theorem 3-11 on the triangle $△ABC$ ,

  $\begin{array}{c}m\angle A+m\angle B+m\angle C=180\\ m\angle A+90+60=180\\ m\angle A+150=180\end{array}$

  $\begin{array}{c}m\angle A=180-150\\ m\angle A=30\end{array}$

Thus, the given triangle is a $30°-60°-90°$ . Here f is the side opposite to the right angle, so it is the hypotenuse and d is the shorter leg and e are the longer leg.

Now, by theorem 8-7, the longer leg is $\sqrt{3}$ times as long as the shorter leg.

Thus,

  $\begin{array}{c}e=\sqrt{3}d\\ 6=\sqrt{3}d\\ \frac{6}{\sqrt{3}}=d\\ 2\sqrt{3}=d\end{array}$

Also, by theorem 8-7, the hypotenuse is two times as long as the shorter side, so

  $\begin{array}{c}f=2d\\ f=2\left(2\sqrt{3}\right)\\ f=4\sqrt{3}\end{array}$

Thus, the complete table is:

d	e	f
$2\sqrt{3}$	$6$	$4\sqrt{3}$