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): Distance \( AB \)

Step 1: Identify the Law of Cosines

We have a triangle \( ABC \) with two sides \( AC = 20 \, \text{km} \), \( BC = 15 \, \text{km} \), and the included angle \( \angle ACB = 110^\circ \). The Law of Cosines states that for a triangle with sides \( a \), \( b \), \( c \) and opposite angles \( A \), \( B \), \( C \) respectively, \( c^2 = a^2 + b^2 - 2ab \cos C \). Here, we want to find \( AB \), let \( AB = c \), \( AC = b = 20 \), \( BC = a = 15 \), and \( C = 110^\circ \).

Step 2: Apply the Law of Cosines

\[

$$\begin{align*} AB^2&=AC^2 + BC^2 - 2 \cdot AC \cdot BC \cdot \cos(\angle ACB)\\ AB^2&=20^2 + 15^2 - 2 \cdot 20 \cdot 15 \cdot \cos(110^\circ) \end{align*}$$

\]
First, calculate \( 20^2 = 400 \), \( 15^2 = 225 \), and \( \cos(110^\circ)\approx - 0.3420 \)
\[

$$\begin{align*} AB^2&=400 + 225 - 2 \cdot 20 \cdot 15 \cdot (-0.3420)\\ AB^2&=625+ 204\\ AB^2&=829 \end{align*}$$

\]

Step 3: Find \( AB \)

Take the square root of \( AB^2 \) to get \( AB \)
\[
AB=\sqrt{829}\approx 28.8 \, \text{km}
\]

Part (b): Shortest distance from \( C \) to \( AB \)

Step 1: Recall the formula for area of a triangle

The area of a triangle can be calculated in two ways: \( \text{Area}=\frac{1}{2}ab\sin C \) (using two sides and included angle) and \( \text{Area}=\frac{1}{2} \times \text{base} \times \text{height} \) (where height is the shortest distance from a vertex to the opposite side). Let the shortest distance from \( C \) to \( AB \) be \( h \) (height), then \( \text{Area}=\frac{1}{2} \times AB \times h \)

Step 2: Calculate the area using \( \frac{1}{2}ab\sin C \)

We know \( a = 15 \), \( b = 20 \), \( C = 110^\circ \), \( \sin(110^\circ)\approx 0.9397 \)
\[
\text{Area}=\frac{1}{2} \times 20 \times 15 \times \sin(110^\circ)
\]
\[
\text{Area}=150\times0.9397 = 140.955 \, \text{km}^2
\]

Step 3: Calculate \( h \) using the other area formula

We know \( \text{Area}=\frac{1}{2} \times AB \times h \), we found \( AB\approx 28.8 \, \text{km} \) and \( \text{Area} = 140.955 \, \text{km}^2 \)
\[
140.955=\frac{1}{2} \times 28.8 \times h
\]
Solve for \( h \):
\[
h=\frac{140.955\times2}{28.8}\approx 9.72 \, \text{km}
\]

Answer:

s:
(a) \( AB\approx \boldsymbol{28.8 \, \text{km}} \) (or more precisely \( \sqrt{829}\approx 28.8 \, \text{km} \))
(b) The shortest distance is approximately \( \boldsymbol{9.72 \, \text{km}} \)