Chapter 14, Problem 16PT

a.

To determine

To find: the height of the equilateral triangle.

Answer to Problem 16PT

The height of the equilateral triangle is $AD=9\sqrt{3}.$

Explanation of Solution

Given information: Suppose a pyramid is built such that a face is an equilateral triangle

of side length 18 feet.

Calculation:

Given the data:

AB=BC=CA= 18

  $AD\perp BC$ .

  

  $CD=BD=\frac{BC}{2}=\frac{18}{2}=9.$

In the right triangle ABD determine AD using the Pythagorean Theorem:

  $\begin{array}{c}A{D}^2+B{D}^2=A{B}^2\\ A{D}^2=A{B}^2-B{D}^2\\ AD=\sqrt{A{B}^2-B{D}^2}=\sqrt{{18}^2-9^2}=\sqrt{9^2\times 3}=9\sqrt{3}\\ AD=9\sqrt{3}.\end{array}$

Therefore, the height of the equilateral triangle is $AD=9\sqrt{3}.$

b.

To determine

To show: that $\sin 2({30}^{\circ })=\sin {60}^{\circ }$ and find the exact values.

Answer to Problem 16PT

  $2\sin {30}^{\circ }.\cos {30}^{\circ }=\sin 2({30}^{\circ })=\frac{\sqrt{3}}{2}.$

Explanation of Solution

Given information: Using the formula

  $\sin 2\theta =2\sin \theta .\cos \theta$ and the measure of the.

equilateral triangle and its height

Calculation:

Given the data:

AB=BC=CA= 18

  $AD\perp BC$ .

  

  $\begin{array}{l}\angle ABD={60}^{\circ }\\ \angle BAD={30}^{\circ }\\ \sin {60}^{\circ }=\sin ABD=\frac{AD}{AB}=\frac{9\sqrt{3}}{18}=\frac{\sqrt{3}}{2}.\end{array}$

  $\begin{array}{l}\sin {30}^{\circ }=\sin BAD=\frac{BD}{AB}=\frac{9}{18}=\frac{1}{2}.\\ \cos {30}^{\circ }=\cos BAD=\frac{AD}{AB}=\frac{9\sqrt{3}}{18}=\frac{\sqrt{3}}{2}.\\ 2\sin {30}^{\circ }.\cos {30}^{\circ }=2\times \frac{1}{2}\times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}=\sin {60}^{\circ }=\sin 2({30}^{\circ })\\ 2\sin {30}^{\circ }.\cos {30}^{\circ }=\sin 2({30}^{\circ })=\frac{\sqrt{3}}{2}.\end{array}$

Therefore, $2\sin {30}^{\circ }.\cos {30}^{\circ }=\sin 2({30}^{\circ })=\frac{\sqrt{3}}{2}.$