Chapter 9.6, Problem 13E

a.

To determine

Tofind:Distance from television

Answer to Problem 13E

  $8\text{ft}$

Explanation of Solution

Given information:

Distance between speakers $=12\text{ ft}$

Concept Used:

In a ${30}^{\circ }-{60}^{\circ }-{90}^{\circ }$ right angled triangle, length of longer leg is product of shorter length and square root of 3 and length of hypotenuse is given by twice the length of shorter leg.

Calculation:

Distance of speaker from television half the distance between both speakers. i.e.

  $\frac{12}{2}\text{ft = }6\text{ ft}$

Now,

Distance of viewer from television (Let $a$ ) will be the longer length of a ${30}^{\circ }-{60}^{\circ }-{90}^{\circ }$ triangle with shorter length of $6\text{ ft}$

Therefore, Distance of viewer from television will be-

  $\begin{array}{l}a=6\sqrt{2}\text{ ft}\\ \Rightarrow a\approx 8\text{ft}\end{array}$

b.

To determine

Tofind:Distance from speaker

Answer to Problem 13E

  $12\text{ ft}$

Explanation of Solution

Given information:

Distance of speakers from television $=6\text{ ft}$

Concept Used:

In a

  ${30}^{\circ }-{60}^{\circ }-{90}^{\circ }$ right angled triangle, length of longer leg is product of shorter length and square root of 3 and length of hypotenuse is given by twice the length of shorter leg.

Calculation:

Distance of viewer from television (Let $b$ ) will be hypotenuse of a ${30}^{\circ }-{60}^{\circ }-{90}^{\circ }$ triangle with shorter length of $6\text{ ft}$

Therefore, Distance of viewer from television will be-

  $\begin{array}{l}b=2\left(6\right)\text{ ft}\\ \Rightarrow b=12\text{ ft}\end{array}$