Back to Exercise: Prove Laws of Sines and Cosines

Exercises: Prove the Laws of Sines and Cosines

Grade 10·21 problems·~35 min·Common Core Math - HS Geometry·standard·hsg-srt-d-10

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Recall and Warm-Up

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

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

$\frac{\sin A}{b} = \frac{\sin B}{a} = \frac{\sin C}{c}$

$\frac{a}{b} = \frac{\sin A}{\sin B} = \frac{c}{\sin C}$

$a \cdot \sin A = b \cdot \sin B = c \cdot \sin C$

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 $h$ ?

$h = \dfrac{b}{\sin A}$

$h = b \sin A$

$h = a \sin A$

$h = b \cos A$

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

$\sin^2\theta + \cos^2\theta = 1$

$\sin^2\theta - \cos^2\theta = 1$

$\sin\theta \cdot \cos\theta = 1$

$\sin\theta + \cos\theta = 1$

Fluency Practice

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: $c^2 =$ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲

$Triangle ABC with angle A = 45 degrees, angle B = 30 degrees, and side a = 12, with side b to be found$

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.

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