Chapter 13.3, Problem 11CYU

a.

To determine

To find:Redraw the diagram so that the angle is in standard position on the coordinate plane.

Answer to Problem 11CYU

Length of $\overline{BC}$ is $24$

Explanation of Solution

Given information:

Angle formed by portable DVD player $=\theta ={125}^0$

Length of the screen $=l=5\frac{1}{2}$ inches.

Diagram of given situation in coordinate plane is

  

b.

To determine

To find: The reference angle.

Answer to Problem 11CYU

Reference angle $={\theta }^{\text{'}}={55}^0$

The trigonometric function that can be used to find the distance to the wall is

  $\cos \left({55}^0\right)=\frac{d}{5\frac{1}{2}}$

Explanation of Solution

Given information:

Angle formed by portable DVD player $=\theta ={125}^0$

Length of the screen $=l=5\frac{1}{2}$ inches.

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

Angle formed by portable DVD player $=\theta ={125}^0$

Length of the screen $=l=5\frac{1}{2}$ inches.

Here $\theta ={125}^0$

It means

The terminal side of ${125}^0$ lies in Quadrant − II

Therefore,

  $\begin{array}{l}{\theta }^{\text{'}}={180}^0-\theta \\ \Rightarrow {\theta }^{\text{'}}={180}^0-{125}^0\\ \Rightarrow {\theta }^{\text{'}}={55}^0\end{array}$

Therefore,

Reference angle $={\theta }^{\text{'}}={55}^0$

We know

  $\cos \left({\theta }^{\text{'}}\right)=\frac{adjacent}{hypotenuse}$

On substituting values

We get

  $\Rightarrow \cos \left({55}^0\right)=\frac{d}{5\frac{1}{2}}$

Therefore,

The trigonometric function that can be used to find the distance to the wall is

  $\cos \left({55}^0\right)=\frac{d}{5\frac{1}{2}}$

c.

To determine

To find: The distance.

Answer to Problem 11CYU

Distance $=d=3.2$ inches.

Explanation of Solution

Given information:

Angle formed by portable DVD player $=\theta ={125}^0$

Length of the screen $=l=5\frac{1}{2}$ inches.

Angle formed by portable DVD player $=\theta ={125}^0$

Length of the screen $=l=5\frac{1}{2}$ inches.

Here $\theta ={125}^0$

It means

The terminal side of ${125}^0$ lies in Quadrant − II

Therefore,

  $\begin{array}{l}{\theta }^{\text{'}}={180}^0-\theta \\ \Rightarrow {\theta }^{\text{'}}={180}^0-{125}^0\\ \Rightarrow {\theta }^{\text{'}}={55}^0\end{array}$

Therefore,

Reference angle $={\theta }^{\text{'}}={55}^0$

We know

  $\cos \left({\theta }^{\text{'}}\right)=\frac{adjacent}{hypotenuse}$

On substituting values

We get

  $\begin{array}{l}\Rightarrow \cos \left({55}^0\right)=\frac{d}{5\frac{1}{2}}\\ \Rightarrow \cos \left({55}^0\right)=\frac{d}{\frac{11}{2}}\\ \Rightarrow d=\frac{11}{2}\cos \left({55}^0\right)\\ \Rightarrow d=\frac{11}{2}\left(0.5736\right)\\ \Rightarrow d=\left(5.5\right)\left(0.5736\right)\\ \Rightarrow d=3.2\end{array}$

Therefore,

Distance $=d=3.2$ inches.