Exercises: Prove Laws of Sines and Cosines - Common Core Math - HS Geometry

A

Recall and Warm-Up

1.

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

A.

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

B.

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

C.

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

D.

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

2.

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 $h$ ?

A.

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

B.

$h = b \sin A$

C.

$h = a \sin A$

D.

$h = b \cos A$

3.

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

A.

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

B.

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

C.

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

D.

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

B

Fluency Practice

1.

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.

2.

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: ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ = ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲

left side:

right side:

3.

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.

4.

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: $c^2 =$ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲

final formula:

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

5.

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.

C

Varied Practice

D

Word Problems

$Top-down diagram of two observation posts A and B, 500 m apart, with angle 65 degrees at A and angle 72 degrees at B toward a wildfire at F, with distance AF to be found$

1.

Two rangers at observation posts $A$ and $B$ , which are 500 m apart, spot a wildfire at point $F$ . Ranger $A$ measures angle $\angle FAB = 65\degree$ and Ranger $B$ measures angle $\angle FBA = 72\degree$ .

Find the distance from post $A$ to the fire. Use $\sin(43\degree) \approx 0.682$ and $\sin(72\degree) \approx 0.951$ . Round to the nearest whole metre.

2.

In triangle $ABC$ , angle $A = 40\degree$ , side $a = 10$ (opposite $A$ ), and side $b = 12$ . Using the Law of Sines: $\sin B = \dfrac{b \sin A}{a} = \dfrac{12 \times 0.643}{10} \approx 0.771$ . Use $\sin^{-1}(0.771) \approx 50.5\degree$ .

1.

How many valid triangles exist with these measurements?

A.

Zero triangles — $\sin B > 1$

B.

Exactly one triangle

C.

Exactly two triangles

D.

The problem cannot be solved without more information

2.

Both values of angle $B$ are $B_1 \approx 50.5\degree$ and $B_2 = 180\degree - 50.5\degree$ . Give the larger value of angle $B$ .

$SAS triangle diagram showing two ships leaving port, one traveling 15 km and the other 20 km with a 70 degree angle between their courses$

3.

Two ships leave port at the same time. Ship A travels 15 km and Ship B travels 20 km. The angle between their courses is 70°.

Find the distance between the two ships. Use $\cos(70\degree) \approx 0.342$ . Round to the nearest tenth.

E

Error Analysis

1.

Priya is given triangle $ABC$ with $a = 8$ , $b = 11$ , and $C = 50\degree$ (the angle between sides $a$ and $b$ ). She writes:
$\frac{8}{\sin A} = \frac{11}{\sin B} = \frac{c}{\sin 50\degree}$
She then tries to solve for $c$ using the Law of Sines.

What error did Priya make, and what should she have done instead?

A.

Priya used the wrong law. With two sides and the included angle (SAS), she should start with the Law of Cosines: $c^2 = 64 + 121 - 2(8)(11)\cos 50\degree$ .

B.

Priya set up the Law of Sines correctly but forgot to cross-multiply to solve for $c$ .

C.

Priya should have placed $\sin(50\degree)$ in the numerator rather than the denominator.

D.

Priya's equation is correct; she just needed to find angles $A$ and $B$ first using $180\degree - A - B$ .

2.

Marcus is solving for side $c$ in triangle $ABC$ with $a = 7$ , $b = 10$ , $C = 60\degree$ . He writes:
$c^2 = 7^2 + 10^2 + 2(7)(10)\cos 60\degree = 49 + 100 + 70 = 219$
$c = \sqrt{219} \approx 14.8$

What error did Marcus make?

A.

Marcus computed $7^2$ and $10^2$ incorrectly.

B.

Marcus wrote $+2ab\cos C$ instead of $-2ab\cos C$ in the Law of Cosines.

C.

Marcus should have used $\cos(30\degree)$ instead of $\cos(60\degree)$ .

D.

Marcus forgot to take the square root of $c^2$ at the end.

F

Challenge

$Triangle ABC with all three sides labeled: side a = 7, side b = 9, and base c = 12, with the largest angle C at the top vertex to be found$

1.

In triangle $ABC$ , all three sides are known: $a = 7$ , $b = 9$ , $c = 12$ . Use the Law of Cosines to find the largest angle $C$ (opposite the longest side $c$ ). Use $\cos^{-1}(-0.111) \approx 96.4\degree$ . Round to the nearest tenth of a degree.

2.