Chapter 6, Problem 7P

Surveying Building Lots A surveyor surveys two adjacent lots and makes the following rough sketch showing his measurements. Calculate all the distances shown in the figure, and use your result to draw an accurate map of the two lots.

To determine

All the distance of the two adjacent lots and draw the accurate map of the two lots.

Answer to Problem 7P

The distance of the two adjacent lots is shown in the Figure (3).

Explanation of Solution

Given:

The figure of the two adjacent lots with measurements are given as,

Figure (1)

Calculation:

From the Figure (1), assume the sides of the building lots be u, v, w, x, y, z.

Figure (2)

Apply angle sum property in triangle ADE to calculate the value of $\angle DAE$,

$\begin{array}{c}\angle ADE+\angle DAE+\angle AED=180\\ 91+\angle DAE+60=180\left(\angle ADE=91°\angle AED=60\right)\\ 151+\angle DAE=180\end{array}$

Solve for the value of $\angle DAE$,

$\begin{array}{c}\angle DAE=180-151\\ =29°\end{array}$

The formula for sine law is given below,

$\frac{a}{\sin A}=\frac{b}{\sin B}$

Substitute 150 for a, x for b, 60 for A, and 29 for B to find the value of x.

$\begin{array}{c}\frac{150}{\sin 60}=\frac{x}{\sin 29}\\ 150\cdot \sin 29=x\sin 60\\ x=\frac{150\cdot \sin 29}{\sin 60}\end{array}$

Further simplify the value,

$\begin{array}{c}x=\frac{\left(150\right)\left(0.484\right)}{0.866}\left(\begin{array}{l}\sin 29°=0.484\\ \sin 60°=0.866\end{array}\right)\\ =83.833\end{array}$

The formula for sine law is given below,

$\frac{a}{\sin A}=\frac{b}{\sin B}$

Substitute 150 for a, y for b, 60 for A, and 91 for B to find the value of y.

$\begin{array}{c}\frac{150}{\sin 60}=\frac{y}{\sin 91}\\ 150\cdot \sin 91=x\cdot \sin 60\\ y=\frac{150\cdot \sin 91}{\sin 60}\end{array}$

Solve for the value of y,

$\begin{array}{c}y=\frac{\left(150\right)\left(0.999\right)}{0.866}\\ =173.036\text{ft}\end{array}$

Apply angle sum property in triangle FAE and calculate the value of $\angle FAE$,

$\begin{array}{c}\angle AFE+\angle FAE+\angle AEF=180\\ 87+\angle FAE+32=180\left(\begin{array}{l}\angle AFE=87\\ \angle AEF=32\end{array}\right)\\ 119+\angle FAE=180\end{array}$

Solve for the value of $\angle FAE$,

$\begin{array}{c}\angle FAE=180-119\\ =61°\end{array}$

The formula for sine law is given below,

$\frac{a}{\sin A}=\frac{b}{\sin B}$

Substitute 173.18 for a, z for b, 87 for A, and 61 for B to find the value of z.

$\begin{array}{c}\frac{173.18}{\sin 87}=\frac{z}{\sin 61}\\ 173.18\cdot \sin 61=z\cdot \sin 87\\ z=\frac{173.18\cdot \sin 61}{\sin 87}\end{array}$

Further solve the value,

$\begin{array}{c}z=\frac{173.18\cdot \left(0.8746\right)}{0.998}\left(\begin{array}{l}\sin 61°=0.8746\\ \sin 87°=0.998\end{array}\right)\\ =151.679\end{array}$

The formula for sine law is given below,

$\frac{a}{\sin A}=\frac{b}{\sin B}$

Substitute 173.18 for a, r for b, 87 for A, and 32 for B to find the value of r.

$\begin{array}{c}\frac{173.18}{\sin 87}=\frac{r}{\sin 32}\left[\frac{a}{\sin A}=\frac{b}{\sin B}\right]\\ 173.18\cdot \sin 32=r\cdot \sin 87\\ r=\frac{173.18\cdot \sin 32}{\sin 87}\end{array}$

Further solve the value,

$\begin{array}{c}r=\frac{173.18\cdot \left(0.529\right)}{0.998}\left(\begin{array}{l}\sin 32°=0.529\\ \sin 87°=0.998\end{array}\right)\\ =91.795\end{array}$

From Figure (2) take the triangle FBE,

Apply angle sum property to calculate the value of $\angle FBE$.

$\begin{array}{c}\angle BFE+\angle FBE+\angle FEB=180\left[\text{sum}\text{of}\text{angles}\text{of}\text{traingle=180}\right]\\ 88+\angle FBE+41=180\\ 129+\angle FBE=180\end{array}$

Solve the value of $\angle FBE$,

$\begin{array}{c}\angle FBE=180-129\\ =51°\end{array}$

The formula for sine law is given below,

$\frac{a}{\sin A}=\frac{b}{\sin B}$

Substitute 151.67 for a, s for b, 51 for A, and 41 for B to find the value of s.

$\begin{array}{c}\frac{151.67}{\sin 51}=\frac{s}{\sin 41}\\ 151.67\cdot \sin 41=s\cdot \sin 51\\ s=\frac{151.67\cdot \sin 41}{\sin 51}\end{array}$

Further solve the value,

$\begin{array}{c}s=\frac{151.67\cdot \left(0.656\right)}{0.7714}\left(\begin{array}{l}\sin 41°=0.656\\ \sin 51°=0.7714\end{array}\right)\\ =128.98\end{array}$

The formula for sine law is given below,

$\frac{a}{\sin A}=\frac{b}{\sin B}$

Substitute 151.67 for a, w for b, 51 for A, and 88 for B to find the value of w.

$\begin{array}{c}\frac{151.67}{\sin 51}=\frac{w}{\sin 88}\left[\frac{a}{\sin A}=\frac{b}{\sin B}\right]\\ 151.67\cdot \sin 88=w\cdot \sin 51\\ w=\frac{151.67\cdot \sin 88}{\sin 51}\end{array}$

Further solve the value,

$\begin{array}{c}w=\frac{151.67\cdot \left(0.9993\right)}{0.7714}\\ =196.496\end{array}$

From Figure (2) take the triangle BCE,

Apply angle sum property to calculate the value of $\angle CBE$,

$\begin{array}{c}\angle BEC+\angle CBE+\angle ECB=180\\ 50+\angle CBE+92=180\\ 142+\angle CBE=180\end{array}$

Solve the value of $\angle CBE$,

$\begin{array}{c}\angle FBE=180-142\\ =38°\end{array}$

The formula for sine law is given below,

$\frac{a}{\sin A}=\frac{b}{\sin B}$

Substitute 195.04 for a, v for b, 92 for A, and 38 for B to find the value of v.

$\begin{array}{c}\frac{195.04}{\sin 92}=\frac{v}{\sin 38}\\ 195.04\cdot \sin 38=v\cdot \sin 92\\ v=\frac{195.04\cdot \sin 38}{\sin 92}\end{array}$

Further solve the value,

$\begin{array}{c}v=\frac{195.04\cdot \left(0.615\right)}{0.999}\\ =121.161\end{array}$

The formula for sine law is given below,

$\frac{a}{\sin A}=\frac{b}{\sin B}$

Substitute 195.04 for a, u for b, 92 for A, and 50 for B to find the value of u.

$\begin{array}{c}\frac{195.04}{\sin 92}=\frac{u}{\sin 50}\left[\frac{a}{\sin A}=\frac{b}{\sin B}\right]\\ 195.04\cdot \sin 50=u\cdot \sin 92\\ u=\frac{195.04\cdot \sin 50}{\sin 92}\end{array}$

Further solve the value,

$\begin{array}{c}u=\frac{195.04\cdot \left(0.766\right)}{0.999}\\ =149.05\end{array}$

Thus, put the value of length on the map,

Figure (3)

Thus, the distances shown in the figure are shown in the above map.