Question

fall 2025 geometry b wwva solving for side lengths of right triangles the measure of angle a is 15°, and the length of side bc is 8. what are the lengths of the other two sides, rounded to the nearest tenth? ac = ab = 2.1 20.7 29.9 40.3

Explanation:

Step1: Use sine function for AB

We know that $\sin(A)=\frac{BC}{AB}$. Given $A = 15^{\circ}$ and $BC = 8$. So $AB=\frac{BC}{\sin(A)}=\frac{8}{\sin(15^{\circ})}$. Since $\sin(15^{\circ})=\sin(45^{\circ}- 30^{\circ})=\sin45^{\circ}\cos30^{\circ}-\cos45^{\circ}\sin30^{\circ}=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}\approx0.259$. Then $AB=\frac{8}{0.259}\approx30.9$.

Step2: Use tangent function for AC

We know that $\tan(A)=\frac{BC}{AC}$. So $AC = \frac{BC}{\tan(A)}$. Since $\tan(15^{\circ})=\tan(45^{\circ}-30^{\circ})=\frac{\tan45^{\circ}-\tan30^{\circ}}{1 + \tan45^{\circ}\tan30^{\circ}}=\frac{1-\frac{\sqrt{3}}{3}}{1 + 1\times\frac{\sqrt{3}}{3}}=2-\sqrt{3}\approx0.268$. Then $AC=\frac{8}{0.268}\approx29.9$.

Answer:

$AC = 29.9$, $AB = 30.9$