Chapter 75, Problem 12A

To determine

(a)

Sketch the rectangular solid for the given views.

Explanation of Solution

The below figure represents the sketch of the sketch of a pyramid formed on surface of the extended sides of the rectangular solid surface block and identified angle.

Figure-(1)

To determine

(b)

The angle of rotation $\angle R$.

Answer to Problem 12A

The angle of rotation $\angle R$ is $49°38^{\prime }$.

Explanation of Solution

The below figure represent the sketch of the sketch of a pyramid formed on surface of the extended sides of the rectangular solid surface block and identified angle.

Figure-(1)

Here the side $DC$ is the common side of the triangle $\Delta ADC$ and $\Delta BDC$.

Consider, the angle $\angle A$ of the triangle $\Delta ADC$ is $41°30^{\prime }$, the angle $\angle B$ of the triangle $\Delta BDC$ is $46°$, the tilt angle of the triangle $\Delta CDE$ is $\angle T$ and side $DC$ is $1\text{unit}$.

Write the expression for the angle $\angle B$ using the triangle $\Delta BDC$.

  $\tan \angle B=\frac{DC}{DB}$ ...... (I)

Here, the length of the common side $DC$ is $DC$ and the length of the side $DB$ is $DB$.

Write the expression of the angle $\angle A$ using the triangle $\Delta ADC$.

  $\tan \angle A=\frac{DC}{AD}$ ........ (II)

Here, the length of the side $AD$ is $AD$.

Consider, the triangle $\Delta ADB$ is a right angle triangle.

Write the expression of the angle $\angle ADB$ of the triangle $\Delta ADB$.

  $\tan \left(\angle ADB\right)=\frac{DB}{AD}$ ......... (III)

Consider, the triangle $\Delta AED$ is a right angle triangle.

Write the expression of the rotation angle.

  $\angle R=90-\angle ADB$ ........ (IV)

Calculation:

Substitute $46°$ for $\angle B$ and $1\text{unit}$ for $DC$ in Equation (I).

  $\begin{array}{l}\tan \left(46°\right)=\frac{\left(1\text{unit}\right)}{DB}\\ DB=\frac{\left(1\text{unit}\right)}{\tan \left(46°\right)}\\ DB=0.97\text{unit}\end{array}$

Substitute $41°30^{\prime }$ for $\angle A$ and $1\text{unit}$ for $DC$ in Equation (II).

  $\begin{array}{l}\tan \left(41°30^{\prime }\right)=\frac{\left(1\text{unit}\right)}{AD}\\ AD=\frac{\left(1\text{unit}\right)}{\tan \left(41.5°\right)}\\ AD=1.14\text{unit}\end{array}$

Substitute $0.97\text{unit}$ for $DB$ and $1.14\text{unit}$ for $AD$ in Equation (III).

  $\begin{array}{l}\tan \left(\angle ADB\right)=\frac{\left(0.97\text{unit}\right)}{\left(1.14\text{unit}\right)}\\ \angle ADB={\tan }^{-1}\left(0.85\right)\\ \angle ADB=40.36°\end{array}$

Substitute $40.36°$ for $\angle ADB$ in Equation (IV).

  $\begin{array}{c}\angle R=90-40.36°\\ =49.64°\\ =49°38^{\prime }\end{array}$

Conclusion:

The angle of rotation $\angle R$ is $49°38^{\prime }$.

To determine

(c)

The angle of tilt $\angle T$.

Answer to Problem 12A

The angle of tilt $\angle T$ is $53°28^{\prime }$.

Explanation of Solution

Write the expression of angle $\angle ADB$ of the triangle $\Delta DCE$.

  $\sin \left(\angle ADE\right)=\frac{DE}{AD}$ ........ (V)

Consider, the triangle $\Delta CDE$ is a right angle triangle.

Write the expression of tilt angle $\angle T$ by using triangle $\Delta CDE$.

  $\tan \angle T=\frac{DC}{DE}$ ........ (VI)

Calculation:

Substitute $40.36°$ for $\angle ADB$ and $1.14\text{unit}$ for $AD$ in Equation (V).

  $\begin{array}{l}\sin \left(40.36°\right)=\frac{DE}{\left(1.14\text{unit}\right)}\\ DE=\left(1.14\text{unit}\right)\sin \left(40.36°\right)\\ DE=0.74\text{unit}\end{array}$

Substitute $1\text{unit}$ for $DC$ and $0.74\text{unit}$ for $DE$ in Equation (VI).

  $\begin{array}{l}\tan \angle T=\frac{\left(1\text{unit}\right)}{\left(0.74\text{unit}\right)}\\ \angle T={\tan }^{-1}\left(1.35\right)\\ \angle T=53.47°\\ \angle T=53°28^{\prime }\end{array}$

Conclusion:

The angle of tilt $\angle T$ is $53°28^{\prime }$.