Exercises: Prove the Laws of Sines and Cosines

Which equation correctly states the Law of Sines for triangle $ABC$ with sides $a$, $b$, $c$ opposite angles $A$, $B$, $C$?

In triangle $ABC$, an altitude $h$ is drawn from vertex $C$ to side $AB$. In the resulting right triangle, $\sin(A) = \dfrac{h}{b}$. What does this give for $h$?

Which of the following is the Pythagorean identity used in the Law of Cosines proof?

In triangle $ABC$, angle $A = 30\degree$, angle $B = 50\degree$, and side $a = 10$. Use $\sin(30\degree) = 0.5$ and $\sin(50\degree) \approx 0.766$. Find side $b$. Round to the nearest tenth.

Complete the key step in the Law of Sines proof. In triangle $ABC$, altitude $h$ from $C$ gives $h = b \sin A$ and $h = a \sin B$. Setting them equal: $b \sin A = a \sin B$. After dividing both sides by $\sin A \cdot \sin B$, the result is:   ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲   =   ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲  

In triangle $ABC$, sides $a = 7$, $b = 10$, and included angle $C = 60\degree$. Use $\cos(60\degree) = 0.5$. Find side $c$. Round to the nearest tenth.

Complete the final step in the Law of Cosines proof. After expanding the distance formula and regrouping, we have: $c^2 = a^2(\cos^2 C + \sin^2 C) + b^2 - 2ab\cos C$. Applying the Pythagorean identity $\sin^2 C + \cos^2 C = 1$ gives: $c^2 =$   ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲  

Triangle $ABC$ has angle $A = 45\degree$, angle $B = 30\degree$, and side $a = 12$. Use $\sin(45\degree) \approx 0.707$ and $\sin(30\degree) = 0.5$. Find side $b$. Round to the nearest tenth.