Exercises: Apply Law of Sines and Cosines - Common Core Math - HS Geometry

A

Recall / Warm-Up

1.

Which formula correctly states the Law of Sines for triangle $ABC$ ?

A.

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

B.

$a^2 = b^2 + c^2 - 2bc\cos A$

C.

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

D.

$a = b\sin A + c\cos A$

2.

A triangle has two sides and the included angle known (SAS). Which law should you use first to find the missing side?

A.

Law of Sines, because you have two ratios to compare

B.

Law of Cosines, because the known angle is between the two known sides

C.

Pythagorean Theorem, because any triangle can be split into right triangles

D.

Either law — they always give the same answer

3.

A bearing of 150° points in which direction relative to north?

A.

60° west of south

B.

30° east of south

C.

150° west of north

D.

30° west of north

B

Fluency Practice

$Triangle ABC with angle A = 35 degrees, angle B = 85 degrees, and side a = 10, with side b to be found$

1.

In triangle $ABC$ , angle $A = 35\degree$ , angle $B = 85\degree$ , and side $a = 10$ . Use $\sin(35\degree) \approx 0.574$ and $\sin(85\degree) \approx 0.996$ . Find side $b$ . Round to the nearest tenth.

2.

In triangle $ABC$ , angle $A = 50\degree$ , angle $C = 70\degree$ , and side $c = 20$ . Use $\sin(50\degree) \approx 0.766$ and $\sin(70\degree) \approx 0.940$ . Find side $a$ . Round to the nearest tenth.

3.

In triangle $ABC$ , angle $A = 30\degree$ , side $a = 8$ , and side $b = 12$ . Compute $\sin B = \dfrac{b \sin A}{a} = \dfrac{12 \times 0.5}{8} = 0.75$ . How many valid triangles exist with these measurements?

A.

No valid triangles — $\sin B > 1$

B.

Exactly one triangle — the acute triangle only

C.

Exactly two triangles — $B \approx 48.6\degree$ or $B \approx 131.4\degree$

D.

Exactly one triangle — the obtuse triangle only

$Triangle ABC with sides a = 8 and b = 11 and included angle C = 50 degrees, with side c to be found$

4.

In triangle $ABC$ , side $a = 8$ , side $b = 11$ , and angle $C = 50\degree$ . Use $\cos(50\degree) \approx 0.643$ . Find side $c$ . Round to the nearest tenth.

5.

In triangle $ABC$ , all three sides are known: $a = 5$ , $b = 7$ , $c = 9$ . Use the Law of Cosines to find angle $C$ (the largest angle, opposite side $c$ ). Use $\cos^{-1}(-0.100) \approx 95.7\degree$ . Round to the nearest tenth of a degree.

C

Varied Practice

1.

In triangle $ABC$ , angle $A = 65\degree$ , angle $B = 75\degree$ , and side $c = 10$ (opposite angle $C$ ). Use $\sin(40\degree) \approx 0.643$ and $\sin(65\degree) \approx 0.906$ . Find side $a$ . Round to the nearest tenth.

$Triangle PQR with sides p = 6 and q = 10 and included angle R = 120 degrees, with side r to be found$

2.

In triangle $PQR$ , side $p = 6$ , side $q = 10$ , and angle $R = 120\degree$ . Use $\cos(120\degree) = -0.5$ . Find side $r$ . Round to the nearest whole number.

$Tip-to-tail vector diagram showing Force 1 of 50 N east and Force 2 of 80 N at 60 degrees, with resultant R to be found$

3.

Two forces act on an object. Force 1 is 50 N directed east (0°). Force 2 is 80 N directed at 60° north of east. The angle between the two force vectors is 60°.

1.

Find the magnitude of the resultant force $R$ . Use $\cos(120\degree) = -0.5$ . Round to the nearest whole number.

2.

Find the angle $\theta$ that the resultant makes with Force 1 (east). Use $\sin^{-1}(0.610) \approx 37.6\degree$ . Round to the nearest tenth.

4.

In triangle $ABC$ , side $b = 15$ , side $c = 12$ , and angle $A = 42\degree$ . Use $\cos(42\degree) \approx 0.743$ . Find side $a$ . Round to the nearest tenth.

5.

In triangle $XYZ$ , angle $X = 40\degree$ , side $x = 10$ (opposite $X$ ), and side $y = 14$ (adjacent to $X$ ). A student finds $\sin Y = \dfrac{14 \sin 40\degree}{10} = 0.900$ , gets $Y \approx 64.2\degree$ , and concludes there is exactly one triangle. What should the student check next?

A.

Whether $\sin Y > 1$ — if not, there is definitely only one triangle

B.

Whether $180\degree - 64.2\degree = 115.8\degree$ also gives a valid triangle when combined with angle $X$

C.

Whether the Law of Cosines gives the same answer as the Law of Sines

D.

Nothing — the student is done; one sine value means one triangle

D

Word Problems

$Top-down surveying diagram showing baseline AB of 80 m with angles 65 degrees at A and 60 degrees at B to a tree at point C$

1.

A surveyor wants to find the width of a canyon. From point $A$ on one side, she measures the angle to a tree at point $C$ on the opposite side as 65°. She then walks 80 m along the canyon edge to point $B$ and measures the angle to the same tree as 60°.

Find the distance from point $A$ to the tree at $C$ . Use $\sin(55\degree) \approx 0.819$ and $\sin(60\degree) \approx 0.866$ . Round to the nearest metre.

$Navigation diagram showing two ships leaving port, one sailing 40 km on bearing 030 and the other 55 km on bearing 120, with a right angle at port$

2.

Two ships leave the same port at the same time. Ship A sails 40 km on a bearing of 030°. Ship B sails 55 km on a bearing of 120°.

Find the distance between the two ships. Use the fact that the angle between their paths is 90°. Round to the nearest tenth of a kilometre.

$Tip-to-tail force diagram showing 60 kN and 90 kN forces at 75 degrees apart, with resultant to be found$

3.

Two tugboats pull a disabled ship. Tugboat 1 exerts a force of 60 kN in one direction. Tugboat 2 exerts a force of 90 kN at an angle of 75° to Tugboat 1. Use $\cos(105\degree) \approx -0.259$ and $\sin(105\degree) \approx 0.966$ .

1.

Find the magnitude of the resultant force. Round to the nearest whole number in kN.

2.

Find the angle $\theta$ that the resultant makes with Tugboat 1's direction. Use $\sin^{-1}(0.722) \approx 46.2\degree$ . Round to the nearest tenth.

E

Error Analysis

1.

Marcus solves a surveying problem: "Two observers at points $A$ and $B$ , which are 100 m apart, both sight the same tree at point $T$ . Observer $A$ measures angle $TAB = 55\degree$ ; observer $B$ measures angle $TBA = 65\degree$ ."

Without a diagram, Marcus writes: "I have two sides and a non-included angle (SSA), so I use the Law of Sines with sides $AB = 100$ and $AT = ?$ and angle $B = 65\degree$ ."

He then solves for $AT$ using the SSA formula and gets $AT \approx 90$ m.

What is the fundamental error in Marcus's reasoning?

A.

Marcus misidentified the case — this is ASA (two angles and the included side between them), not SSA

B.

Marcus should have added the two angles: $55\degree + 65\degree = 120\degree$ , then used the Law of Cosines

C.

Marcus's arithmetic is wrong — the correct answer is $AT \approx 107$ m, not 90 m

D.

Marcus used the right method; 90 m is a reasonable answer for this problem

2.

Priya is finding an unknown angle using SSA: triangle $ABC$ with $A = 35\degree$ , $a = 6$ , $b = 9$ .

She computes $\sin B = \dfrac{9 \sin 35\degree}{6} \approx 0.861$ , finds $B_1 \approx 59.5\degree$ , and stops, writing: "There is exactly one solution: $B = 59.5\degree$ ."

What is wrong with Priya's conclusion?

A.

Priya's sine calculation is wrong — $\dfrac{9 \times 0.574}{6} \neq 0.861$

B.

Priya should have used the Law of Cosines for an SSA problem

C.

Priya found one valid solution but did not check whether $B_2 = 180\degree - 59.5\degree = 120.5\degree$ also gives a valid triangle

D.

Priya's answer is correct — one sine value always means exactly one triangle

F

Challenge / Extension

$Top-down surveying diagram with baseline AB of 120 m, angle 65 degrees at A and 55 degrees at B, with distance BC to be found$

1.

A surveyor at point $A$ wants to find the distance to a tree at point $C$ across a canyon. She measures a baseline $AB = 120$ m along the canyon edge. From $A$ , she measures the angle to $C$ as $65\degree$ . From $B$ , she measures the angle to $C$ as $55\degree$ . Find the distance $BC$ . Use $\sin(65\degree) \approx 0.906$ and $\sin(60\degree) \approx 0.866$ . Round to the nearest metre.

2.