Question

question 26 of 26 a car travels 1.2 miles from point a to point b. the car then turns at point b and travels 1.8 miles to point c, before heading back to point a. the distance from point c to point a is 1.6 miles. if the car’s path is represented by a triangle, what angle turn, to the nearest tenth of a degree, did the car make at point b?
a. 65.9°
b. 60.6°
c. 73.4°
d. 50.3°

Explanation:

Step1: Identify the triangle sides

We have a triangle \( ABC \) where \( AB = 1.2 \) miles, \( BC = 1.8 \) miles, and \( AC = 1.6 \) miles. We need to find the angle at \( B \), so we can use the Law of Cosines. The Law of Cosines formula for angle \( B \) is:
\[
AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(\angle B)
\]

Step2: Substitute the values

Substitute \( AB = 1.2 \), \( BC = 1.8 \), and \( AC = 1.6 \) into the formula:
\[
1.6^2 = 1.2^2 + 1.8^2 - 2 \cdot 1.2 \cdot 1.8 \cdot \cos(\angle B)
\]
Calculate each term:
\( 1.6^2 = 2.56 \), \( 1.2^2 = 1.44 \), \( 1.8^2 = 3.24 \)
So the equation becomes:
\[
2.56 = 1.44 + 3.24 - 4.32 \cdot \cos(\angle B)
\]

Step3: Simplify the equation

First, add \( 1.44 \) and \( 3.24 \):
\( 1.44 + 3.24 = 4.68 \)
So:
\[
2.56 = 4.68 - 4.32 \cdot \cos(\angle B)
\]
Subtract \( 4.68 \) from both sides:
\[
2.56 - 4.68 = - 4.32 \cdot \cos(\angle B)
\]
\( 2.56 - 4.68 = -2.12 \)
So:
\[
-2.12 = - 4.32 \cdot \cos(\angle B)
\]
Divide both sides by \( -4.32 \):
\[
\cos(\angle B) = \frac{-2.12}{-4.32} \approx 0.4907
\]

Step4: Find the angle

Take the inverse cosine (arccos) of \( 0.4907 \):
\[
\angle B = \arccos(0.4907) \approx 60.6^\circ
\]

Answer:

B. \( 60.6^\circ \)