Solve the problem.
-The length of the base of an isosceles triangle is $13.02$ meters. Each base angle is $28.04 ^ { \circ }$. Find the length of each of the two equal sides of the triangle. Round your answer to the hundredths place.
A) $7.38 \mathrm {~m}$
B) $12.22 \mathrm {~m}$
C) $13.85 \mathrm {~m}$
D) $14.75 \mathrm {~m}$

Question

Quizplus · Accepted Answer

The correct answer is A. The length of each of the two equal sides of the triangle can be found using the Law of Cosines:$$c^2 = a^2 + b^2 - 2ab \cos(C)$$where $c$ is the length of the base, $a$ and $b$ are the lengths of the equal sides, and $C$ is the angle between them.In this case, $c = 13.02$ meters, $C = 28.04 ^ { \circ }$, and $a = b$. Substituting these values into the equation, we get:$$b^2 = \frac{c^2 + b^2 - 2ab \cos(C)}{2}$$$$b^2 = \frac{13.02^2 + b^2 - 2b^2 \cos(28.04 ^ { \circ })}{2}$$$$b^2 = \frac{170.12 + b^2 - 2b^2 (0.8829)}{2}$$$$b^2 = \frac{170.12 + b^2 - 1.7658b^2}{2}$$$$b^2 = \frac{170.12 - 0.7658b^2}{2}$$$$b^2 = 85.06 - 0.3829b^2$$$$1.3829b^2 = 85.06$$$$b^2 = 61.51$$$$b = \sqrt{61.51} = 7.84 \mathrm {~m}$$Therefore, the length of each of the two equal sides of the triangle is $7.84 \mathrm {~m}$.

Solve the Problem 13.0213.0213.02 Meters Each Base Angle Is 28.04∘28.04 ^ { \circ }28.04∘

Solve the Problem $13.02$ Meters Each Base Angle Is $28.04 ^ { \circ }$