Chapter 13.3, Problem 32PPS

a.

To determine

To find: Reference angle.

Answer to Problem 32PPS

Explanation of Solution

Given information:

  $\theta ={154}^0$

Height goalie jumps to catch ball $=h=4$

  $x$ is the distance between soccer player and goalie.

Concept:

If $\theta$ is the non-quadrantal angle in standard position, its reference angle ${\theta }^{\text{'}}$ is the acute angle formed by the terminal side of $\theta$ and the x-axis.

The ruled for finding the measures of reference angles for $0^0<\theta <{360}^0$ or $0^0<\theta <2\pi$ are shown

For Quadrant − I

  ${\theta }^{\text{'}}=\theta$

For Quadrant − II

  ${\theta }^{\text{'}}={180}^0-\theta$ or ${\theta }^{\text{'}}=\pi -\theta$

For Quadrant − III

  ${\theta }^{\text{'}}=\theta -{180}^0$ or ${\theta }^{\text{'}}=\theta -\pi$

For Quadrant − IV

  ${\theta }^{\text{'}}={360}^0-\theta$ or ${\theta }^{\text{'}}=2\pi -\theta$

Given

Here $\theta ={154}^0$

It means

The terminal side of ${154}^0$ lies in Quadrant − II

Therefore,

  $\begin{array}{l}{\theta }^{\text{'}}={180}^0-\theta \\ \Rightarrow {\theta }^{\text{'}}={180}^0-{154}^0\\ \Rightarrow {\theta }^{\text{'}}={26}^0\end{array}$

Therefore,

Reference angle $={\theta }^{\text{'}}={26}^0$

We know

  $\tan \left(x\right)=\frac{opposite}{adjacent}$

Therefore,

On substituting values

We get

  $\tan \left({26}^0\right)=\frac{7}{x}$

b.

To determine

To find: About how far away from the goalie was the soccer player.

Answer to Problem 32PPS

Distance between goalie and soccer player $=x=14.3$ feet

Explanation of Solution

Given information:

  $\theta ={154}^0$

Height goalie jumps to catch ball $=h=4$

  $x$ is the distance between soccer player and goalie.

Given

Here $\theta ={154}^0$

It means

The terminal side of ${154}^0$ lies in Quadrant − II

Therefore,

  $\begin{array}{l}{\theta }^{\text{'}}={180}^0-\theta \\ \Rightarrow {\theta }^{\text{'}}={180}^0-{154}^0\\ \Rightarrow {\theta }^{\text{'}}={26}^0\end{array}$

Therefore,

Reference angle $={\theta }^{\text{'}}={26}^0$

We know

  $\tan \left(x\right)=\frac{opposite}{adjacent}$

Therefore,

On substituting values

We get

  $\begin{array}{l}\tan \left({26}^0\right)=\frac{7}{x}\\ \Rightarrow x=\frac{7}{\tan \left({26}^0\right)}\\ \Rightarrow x=\frac{7}{0.49}\\ \Rightarrow x=14.3\end{array}$

Therefore,

Distance between goalie and soccer player $=x=14.3$ feet