Question

In triangle ABC, side AB has a length of 8 cm and side BC has a length of 5 cm. If the triangle has an area of 10 square centimeters, what is the measure of the angle between sides AB and BC?
135°
150°
120°
45°
30°
60°

Thank you!

Answer 1

Recall the following formula:

Given a triangle with side lengths a, b, and c. Let the measure of the angle between sides of length a and c be B.

Then, the area of the triangle is given by :

                                 $Area= \frac{1}{2}\cdot a\cdot c \cdot \sin B$.

 In our example we have: Area=10 square cm, a=8 cm, c=5 cm, and we want to find the measure of the angle between the 2 sides a and c.

Substituting in the formula we have:

$10= \frac{1}{2}\cdot 8\cdot 5 \cdot \sin B$

Thus, $10=20 \sin B$, which means $\sin B = \frac{1}{2}$.

$\frac{1}{2}$ is the sine of $30^{\circ}$, but also $150^{\circ}$.


Answer: both 30, and 150 degrees are possible