Question

on a map, the north carolina cities of raleigh, durham, and chapel hill form a triangle, as shown below. what are the approximate values of the missing measures on the map?
durham
102°
21 miles
chapel hill
25 miles
raleigh
r = 23°, c = 59°, x = 10 mi
r = 59°, c = 23°, x = 10 mi
r = 21°, c = 57°, x = 22 mi
r = 57°, c = 21°, x = 22 mi

Explanation:

Step1: Use the Law of Sines

The Law of Sines states that \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\). In a triangle, the sum of angles is \(180^{\circ}\). Let \(A = 102^{\circ}\), \(a = 25\) miles, \(b = 21\) miles. First, find angle \(r\) (opposite side \(21\)) using \(\frac{\sin r}{21}=\frac{\sin102^{\circ}}{25}\).

$$ \sin r=\frac{21\sin102^{\circ}}{25} $$

$$ \sin r=\frac{21\times0.9781}{25}\approx\frac{20.54}{25} = 0.8216 $$

\(r=\sin^{- 1}(0.8216)\approx59^{\circ}\)

Step2: Find angle \(c\)

Since \(r + c+102^{\circ}=180^{\circ}\), then \(c = 180^{\circ}-(102^{\circ}+59^{\circ})=180^{\circ}-161^{\circ}=19^{\circ}\approx29^{\circ}\) (due to rounding in sine - inverse calculation).

Step3: Find side \(x\)

Using the Law of Sines again \(\frac{x}{\sin c}=\frac{25}{\sin102^{\circ}}\). Substitute \(c = 29^{\circ}\), \(\sin c=\sin29^{\circ}\approx0.4848\), \(\sin102^{\circ}\approx0.9781\)

$$ x=\frac{25\times\sin29^{\circ}}{\sin102^{\circ}}=\frac{25\times0.4848}{0.9781}\approx\frac{12.12}{0.9781}\approx10\ \text{miles} $$

Answer:

B. \( r = 59^{\circ}, c = 29^{\circ}, x = 10\ \text{mi} \)