Question

12 multiple choice 1 point each of the two equal sides of an isosceles triangle is 66.9 inches long. the base of this triangle is 92.3 inches long. what is the size of the two equal base angles of this triangle? (give your answer in degrees and round your answer to the nearest tenth of a degree.) 46.4° 46.2° 45.6° 46.0° 45.8°

Explanation:

Step1: Use the cosine - law

Let the equal sides be $a = b=66.9$ inches and the base $c = 92.3$ inches. The cosine - law for finding an angle $\theta$ in a triangle is $\cos\theta=\frac{a^{2}+c^{2}-b^{2}}{2ac}$. In an isosceles triangle, if we want to find the base - angle, we can use $\cos\theta=\frac{\frac{c}{2}}{a}$. First, we consider half of the base, $\frac{c}{2}=\frac{92.3}{2}=46.15$ inches.

Step2: Calculate the cosine of the angle

$\cos\theta=\frac{46.15}{66.9}\approx0.69$.

Step3: Find the angle

$\theta=\cos^{- 1}(0.69)\approx46.4^{\circ}$.

Answer:

$46.4^{\circ}$