Chapter 76, Problem 13A

To determine

(a)

The sketch and label of a rectangular solid and the pyramid formed by the angular surface edges.

Answer to Problem 13A

The sketch and label of a rectangular solid and the pyramid formed by the angular surface edges is

  

Explanation of Solution

Consider the angle $\angle A=37°.20^{\prime }$ of triangle $ADE$, the angle $\angle B=71°5^{\prime }$ of triangle $BDE$.

There are mainly three types of views such as top view, front view and right side view. The rectangular solid and pyramid is formed using the trihedral method. In this method all the planes are perpendicular to each other and eight right triangles are formed. The object is placed in these right angles to take the projection.

The steps of the sketching of the rectangular solid and the pyramid formed by the surface to be cut and the extended sides of the block are following.

1.Draw a rectangular solid surface block.

2.Draw triangles on the extended surface of the rectangular solid surface block.

3.Cut the extended surface of the rectangular solid surface block as per the drawn triangle on the surface of the block to form a pyramid on the extended sides of the block.

4.The sketch of a pyramid formed on surface of the extended sides of the rectangular solid surface block.

5.Identify the angle $\angle R$ and $\angle C$ and the given angles in the sketch of a pyramid formed on surface of the extended sides of the rectangular solid surface block.

  

  Figure-(1)

To determine

(b)

The angle $\angle R$.

Answer to Problem 13A

The angle $\angle R$ is $14.24°$.

Explanation of Solution

Write the expression for the angle $\angle B$ using the triangle $BDE$.

  $\tan B=\frac{DE}{BD}$ ...... (I)

Here, the length of the side $DE$ is $DE$ and the length of the side $BD$ is $BD$.

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

  $\tan A=\frac{DE}{AD}$ ...... (II)

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

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

  $\tan R=\frac{BD}{BC}$ ...... (III)

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

Calculation:

Consider the length of the side $DE=1\text{unit}$.

Substitute $1\text{unit}$ for $DE$ and $71°50^{\prime }$ for $B$ in Equation (I).

  $\begin{array}{l}\tan 71°50^{\prime }=\frac{\text{1}\text{unit}}{BD}\\ BD=\frac{\text{1}\text{unit}}{\tan 71°50^{\prime }}\\ BD=\frac{\text{1}\text{unit}}{2.9886}\\ BD=0.3346\text{unit}\end{array}$

Substitute $1\text{unit}$ for $DE$ and $37°20^{\prime }$ for $A$ in Equation (II).

  $\begin{array}{l}\tan 37°20^{\prime }=\frac{\text{1}\text{unit}}{AD}\\ AD=\frac{\text{1}\text{unit}}{\tan 37°20^{\prime }}\\ AD=\frac{\text{1}\text{unit}}{0.7590}\\ AD=1.3175\text{unit}\end{array}$

The length of the side $AD$ and $BC$ are same. Therefore, the length of the side $BC$ is $1.3175\text{unit}$.

Substitute $1.3175\text{unit}$ for $BC$ and $0.3346\text{unit}$ for $BD$ in Equation (III).

  $\begin{array}{l}\tan R=\frac{0.3346\text{unit}}{1.3175\text{unit}}\\ \tan R=0.2539\\ R={\tan }^{-1}\left(0.2539\right)\\ R=14.24°\end{array}$

Conclusion:

The angle $\angle R$ is $14.24°$.

To determine

(c)

The angle $\angle C$.

Answer to Problem 13A

The angle $\angle C$ is $36.32°$.

Explanation of Solution

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

  $\sin R=\frac{BD}{CD}$ ...... (IV)

Write the expression for the angle $\angle C$ using triangle $CDE$.

  $\tan C=\frac{DE}{CD}$ ...... (V)

Calculation:

Consider the length of the side $DE=1\text{unit}$.

Substitute $0.3346\text{unit}$ for $BD$ and $14.24°$ for $R$ in Equation (IV).

  $\begin{array}{l}\sin 14.24°=\frac{0.3346\text{unit}}{CD}\\ CD=\frac{0.3346\text{unit}}{\sin 14.24°}\\ CD=\frac{0.3346\text{unit}}{0.2460}\\ CD=1.3601\text{unit}\end{array}$

Substitute $1.3601\text{unit}$ for $CD$ and $1\text{unit}$ for $DE$ in Equation (V).

  $\begin{array}{l}\tan C=\frac{1\text{unit}}{1.3601\text{unit}}\\ \tan C=0.7352\\ C={\tan }^{-1}\left(0.7352\right)\\ C=36.32°\end{array}$

Conclusion:

The angle $\angle C$ is $36.32°$.