Question

1. in the diagram, a, b, and c represent three towns. they are joined by three straight roads. the distance ac=20km,bc=15km, and $\angle acb = 110^\circ$. calculate: (a) the distance ab. (b) the shortest distance from c to road ab.

Explanation:

Part (a)

Step1: Identify the Law to Use

We have a triangle \(ABC\) with two sides \(AC = 20\space km\), \(BC=15\space km\) and the included angle \(\angle ACB = 110^{\circ}\). To find the side \(AB\), we use the Law of Cosines. The Law of Cosines states that for a triangle with sides \(a\), \(b\), \(c\) and the included angle \(C\) (opposite to side \(c\)):
\[c^{2}=a^{2}+b^{2}-2ab\cos C\]
In triangle \(ABC\), let \(AB = c\), \(AC = b=20\space km\), \(BC=a = 15\space km\) and \(\angle ACB=C = 110^{\circ}\)

Step2: Substitute the Values into the Formula

Substitute \(a = 15\), \(b = 20\) and \(C=110^{\circ}\) into the formula:
\[

$$\begin{align*} c^{2}&=15^{2}+20^{2}-2\times15\times20\times\cos(110^{\circ})\\ c^{2}&=225 + 400-600\times\cos(110^{\circ}) \end{align*}$$

\]
We know that \(\cos(110^{\circ})=\cos(180^{\circ} - 70^{\circ})=-\cos(70^{\circ})\approx - 0.3420\)
\[

$$\begin{align*} c^{2}&=625-600\times(- 0.3420)\\ c^{2}&=625 + 205.2\\ c^{2}&=830.2 \end{align*}$$

\]

Step3: Find the Value of \(c\)

Take the square root of \(c^{2}\) to find \(c\):
\[c=\sqrt{830.2}\approx28.81\space km\]

Part (b)

Step1: Recall the Formula for Area of a Triangle

The area of a triangle can be calculated in two ways:

1. Using two sides and the included angle: \(Area=\frac{1}{2}ab\sin C\)
2. Using base and height: \(Area=\frac{1}{2}\times base\times height\)

Let the shortest distance from \(C\) to \(AB\) be \(h\) (this is the height when \(AB\) is the base). First, we calculate the area using the formula \(\frac{1}{2}ab\sin C\) and then equate it to \(\frac{1}{2}\times AB\times h\) to find \(h\)

Step2: Calculate the Area Using \(\frac{1}{2}ab\sin C\)

We have \(a = 15\), \(b = 20\) and \(C = 110^{\circ}\)
\[

$$\begin{align*} Area&=\frac{1}{2}\times15\times20\times\sin(110^{\circ})\\ &=150\times\sin(110^{\circ}) \end{align*}$$

\]
\(\sin(110^{\circ})=\sin(180^{\circ}- 70^{\circ})=\sin(70^{\circ})\approx0.9397\)
\[

$$\begin{align*} Area&=150\times0.9397\\ &=140.955\space km^{2} \end{align*}$$

\]

Step3: Calculate the Height \(h\)

We know that \(Area=\frac{1}{2}\times AB\times h\). We found \(AB\approx28.81\space km\) and \(Area = 140.955\space km^{2}\)
\[

$$\begin{align*} 140.955&=\frac{1}{2}\times28.81\times h\\ h&=\frac{140.955\times2}{28.81}\\ h&\approx\frac{281.91}{28.81}\\ h&\approx9.78\space km \end{align*}$$

\]

Answer:

s:
(a) The distance \(AB\) is approximately \(\boldsymbol{28.81\space km}\)

(b) The shortest distance from \(C\) to road \(AB\) is approximately \(\boldsymbol{9.78\space km}\)