Question

the length of the base of an isosceles triangle is 30.54 inches. each base angle is 30.72°. find the length of each of the two equal sides of the triangle. (hint: divide the triangle into two right triangles.) the other two sides of the triangle have lengths of inches. (simplify your answer. type an integer or decimal rounded to the nearest hundredth as needed.)

Explanation:

Step1: Divide the isosceles triangle into two right - triangles

Draw an altitude from the vertex angle to the base of the isosceles triangle. This divides the base of length $b = 30.54$ inches into two equal parts, each of length $\frac{b}{2}=\frac{30.54}{2}=15.27$ inches. The base - angle $\theta=30.72^{\circ}$.

Step2: Use the cosine function in the right - triangle

In a right - triangle, $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$. Let the length of the equal sides of the isosceles triangle be $s$ (which is the hypotenuse of the right - triangle), and the adjacent side to the base - angle $\theta$ is $\frac{b}{2}$. So, $\cos\theta=\frac{\frac{b}{2}}{s}$, and we can solve for $s$: $s=\frac{\frac{b}{2}}{\cos\theta}$.

Step3: Substitute the values

Substitute $\frac{b}{2}=15.27$ inches and $\theta = 30.72^{\circ}$ into the formula. $\cos(30.72^{\circ})\approx0.8609$. Then $s=\frac{15.27}{0.8609}\approx17.74$ inches.

Answer:

$17.74$