Chapter 76, Problem 13A

To determine

(a)

The sketch and label of a rectangular solid.

Answer to Problem 13A

The sketch and label of a rectangular is shown in Figure-(1).

Explanation of Solution

Project auxiliary lines that form right triangles containing the given angles, the axis of the hole, and the angles to be computed, $\angle R$ and $\angle T$. $BC$ is a side of $\Delta BCD$, which contains the given $19°00^{\prime }$ angle. $BC$ is also a side of right $\Delta BCE$, which contains $\angle R$ that is to be computed.

The sketch and label of a rectangular is shown below.

Figure-(1)

To determine

(b)

The angle of rotation $\angle R$.

Answer to Problem 13A

The angle of rotation $\angle R$ is $30°49^{\prime }$.

Explanation of Solution

Write the expression for the angle $\angle D$ using the triangle $BCD$.

  $\tan D=\frac{BC}{DC}$ ...... (I)

Here, the length of the side $BC$ is $BC$ and the length of the side $DC$ is $DC$.

Write the expression for the angle $\angle A$ using triangle $ACE$.

  $\tan A=\frac{CE}{AE}$ ...... (II)

Here, the length of the side $CE$ is $CE$ and the length of the side $AE$ is $AE$.

Write the expression for the angle $\angle R$ using triangle $BCE$.

  $\tan \angle R=\frac{BC}{CE}$ ...... (III)

Calculation:

Consider the length of the side $BC$ is equal to $1\text{unit}$.

Substitute $1\text{unit}$ for $BC$ and $\tan 19°00^{\prime }$ for $\tan D$ in Equation (I).

  $\begin{array}{l}\tan 19°00^{\prime }=\frac{\text{1}\text{unit}}{DC}\\ DC=\frac{\text{1}\text{unit}}{\tan 19°00^{\prime }}\\ DC=\frac{\text{1}\text{unit}}{0.3443}\\ DC=2.9042\text{unit}\end{array}$

The length of the side $AE$ and $DC$ are same. Therefore, the length of the side $AE$ is $2.9042\text{unit}$.

  $AE=2.9042\text{unit}$

Substitute $2.9042\text{unit}$ for $AE$ and $30°00^{\prime }$ for $A$ in Equation (II).

  $\begin{array}{l}\tan 30°00^{\prime }=\frac{CE}{2.9042\text{unit}}\\ CE=\left(2.9042\text{unit}\right)\tan 30°00^{\prime }\\ CE=\left(2.9042\text{unit}\right)\left(0.5773\right)\\ CE=1.6767\text{unit}\end{array}$

Substitute $1.6767\text{unit}$ for $CE$ and $1\text{unit}$ for $BC$ in Equation (III).

  $\begin{array}{l}\tan \angle R=\frac{1\text{unit}}{1.6767\text{unit}}\\ \tan \angle R=0.5964\\ \angle R={\tan }^{-1}\left(0.5964\right)\\ \angle R=30°49^{\prime }\end{array}$

Conclusion:

The angle $\angle R$ is $30°49^{\prime }$.

To determine

(c)

The angle of tilt $\angle C$.

Answer to Problem 13A

The angle of tilt $\angle T$ is $33°54^{\prime }$.

Explanation of Solution

Write the expression for the length $BE$ using the triangle $BCE$.

  $BE=\sqrt{B{C}^2+C{E}^2}$ ...... (IV)

Write the expression for the angle $\angle T$ using triangle $AEB$.

  $\tan \angle T=\frac{BE}{AE}$ ...... (V)

Calculation:

Substitute $1\text{unit}$ for $BC$ and $1.6767\text{unit}$ for $CE$ in Equation (IV).

  $\begin{array}{c}BE=\sqrt{{\left(1\text{unit}\right)}^2+{\left(1.6767\text{unit}\right)}^2}\\ =\sqrt{{\left(3.8113\text{unit}\right)}^2}\\ =1.9522\text{unit}\end{array}$

Substitute $1.9522\text{unit}$ for $BE$ and $2.9042\text{unit}$ for $AE$ in Equation (V).

  $\begin{array}{l}\tan \angle T=\frac{1.9522\text{unit}}{2.9042\text{unit}}\\ \tan \angle T=0.6721\\ \angle T={\tan }^{-1}\left(0.6721\right)\\ \angle T=33°54^{\prime }\end{array}$

Conclusion:

The angle $\angle T$ is $33°54^{\prime }$.