Triangle Sides 73C1Ee

1. **State the problem:** We have a right triangle $\triangle ABC$ with a right angle at $B$, angle $A = 60^\circ$, and side $AB = 13$. We need to find the lengths of $AC$ (the hypotenuse) and $BC$ (the other leg). 2. **Identify the sides:** In right triangle $ABC$, $AB$ and $BC$ are legs, and $AC$ is the hypotenuse opposite the right angle. 3. **Use trigonometric ratios:** Since $\angle A = 60^\circ$, side $AB$ is adjacent to $\angle A$, and side $BC$ is opposite $\angle A$. 4. **Formulas:** - $\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}$ - $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$ 5. **Calculate $AC$ using cosine:** $$\cos(60^\circ) = \frac{AB}{AC}$$ $$\cos(60^\circ) = \frac{13}{AC}$$ Since $\cos(60^\circ) = \frac{1}{2}$, $$\frac{1}{2} = \frac{13}{AC}$$ Multiply both sides by $AC$: $$\frac{1}{2} AC = 13$$ Divide both sides by $\frac{1}{2}$: $$AC = \frac{13}{\frac{1}{2}} = 13 \times 2 = 26$$ 6. **Calculate $BC$ using sine:** $$\sin(60^\circ) = \frac{BC}{AC}$$ $$\sin(60^\circ) = \frac{BC}{26}$$ Since $\sin(60^\circ) = \frac{\sqrt{3}}{2}$, $$\frac{\sqrt{3}}{2} = \frac{BC}{26}$$ Multiply both sides by $26$: $$BC = 26 \times \frac{\sqrt{3}}{2} = 13 \sqrt{3}$$ 7. **Final answers:** - $AC = 26$ - $BC = 13 \sqrt{3}$