Question

question 25 of 26
a ship travels 10 miles from point a to point b, makes a turn of 106°, and travels 14 miles to point c. if the ship travels directly from point c back to point a, how many miles will it travel on the last leg of the trip (from point c to point a)? round your answer to the nearest tenth of a mile.

a. 23.1 miles

b. 21.5 miles

c. 19.3 miles

d. 24.8 miles

Explanation:

Step1: Identify the Law to Use

We can use the Law of Cosines here. The Law of Cosines states that for a triangle with sides \(a\), \(b\), \(c\) and the angle \(C\) opposite side \(c\), \(c^{2}=a^{2}+b^{2}-2ab\cos(C)\). In triangle \(ABC\), \(AB = 10\) miles, \(BC=14\) miles, and the included angle \(\angle ABC = 106^{\circ}\). We need to find the length of \(AC\). Let \(a = 14\), \(b = 10\), and \(C=106^{\circ}\), then \(AC^{2}=a^{2}+b^{2}-2ab\cos(C)\).

Step2: Substitute the Values

Substitute \(a = 14\), \(b = 10\), and \(C = 106^{\circ}\) into the formula. First, calculate \(\cos(106^{\circ})\). Using a calculator, \(\cos(106^{\circ})\approx - 0.2756\). Then:
\[

$$\begin{align*} AC^{2}&=14^{2}+10^{2}-2\times14\times10\times\cos(106^{\circ})\\ &=196 + 100-280\times(- 0.2756)\\ &=296+77.168\\ &=373.168 \end{align*}$$

\]

Step3: Find the Length of \(AC\)

Take the square root of \(AC^{2}\) to find \(AC\):
\[
AC=\sqrt{373.168}\approx19.3
\]

Answer:

C. 19.3 miles