Back to Law of Sines

Law of Sines

Grade 10·20 problems·~30 min·ACT Math·topic·act-geo-trig-lawsines
Work through problems with immediate feedback
A

Recall / Warm-Up

1.

In a right triangle, if the side opposite an angle is 5 and the hypotenuse is 13, what is the sine of that angle?

2.

If sin(θ)=0.6\sin(\theta) = 0.6, what is θ\theta to the nearest degree?

3.

In triangle ABC, angle A = 50° and angle B = 70°. What is angle C in degrees?

B

Fluency Practice

1.

In triangle ABC, angle A = 40°, angle B = 75°, and side a=10a = 10. Find side bb. Round to the nearest hundredth.

2.

In triangle DEF, angle D = 35°, angle F = 80°, and side f=20f = 20. Find side dd. Round to the nearest hundredth.

3.

In triangle PQR, angle P = 50°, angle Q = 60°, and side p=14p = 14. What is side qq? Round to the nearest tenth.

4.

In triangle ABC, angle A = 30°, side a=8a = 8, and side b=10b = 10. Find angle B in degrees. Round to the nearest tenth.

5.

In triangle XYZ, angle X = 45°, angle Y = 100°, and side y=25y = 25. Find side xx. Round to the nearest tenth.

C

Varied Practice

1.

A triangle has the following known information: two angles and the side between them. Which case is this, and can the Law of Sines be used?

2.

To find side bb in triangle ABC using the Law of Sines with a=12a = 12, angle A=42°A = 42\degree, and angle B=73°B = 73\degree:

Step 1: The Law of Sines states asinA=bsinB\frac{a}{\sin A} = \frac{b}{\sin B}.

Step 2: Substitute: 12sin42°=bsin73°\frac{12}{\sin 42\degree} = \frac{b}{\sin 73\degree}.

Step 3: Solve for bb: b=12×sin°sin°b = \frac{12 \times \sin \underline{\hspace{5em}}\degree}{\sin \underline{\hspace{5em}}\degree}.

numerator angle:
denominator angle:
3.

You are given triangle information: side a = 9, side b = 14, and angle C = 55°. Which method should you use to solve this triangle?

4.

In triangle ABC, angle A = 30°, side a=7a = 7, and side b=14b = 14. You set up sinB=14×sin30°7=1.0\sin B = \frac{14 \times \sin 30\degree}{7} = 1.0. How many triangles can be formed?

5.

In triangle ABC, angle A = 40°, side a=10a = 10, and side b=15b = 15. You compute sinB=15×sin40°100.964\sin B = \frac{15 \times \sin 40\degree}{10} \approx 0.964. Angle B74.6°B \approx 74.6\degree. Does a second triangle exist?

D

Word Problems / Application

1.

Two fire lookout towers are 30 miles apart. Tower A spots a fire at an angle of 48° from the line connecting the two towers. Tower B spots the same fire at an angle of 63° from the same line.

How far is the fire from Tower A? Round to the nearest tenth of a mile.

2.

A surveyor stands at point P and measures the angle to two landmarks, Q and R. The angle QPR = 35°. She knows that PR = 200 m and QR = 120 m. She wants to find angle PQR using the Law of Sines.

How many possible triangles can be formed with this information?

3.

A boat leaves port and sails 18 km on a course that makes a 52° angle with the shoreline. The boat then turns and heads toward a lighthouse that is down the coast. From the lighthouse, the angle between the shoreline and the line of sight to the port is 44°.

How far is the boat from the lighthouse after it turns? Round to the nearest tenth of a kilometer.

E

Error Analysis

1.

Marcus solved this problem:

"In triangle ABC, angle A = 50°, angle B = 65°, and side b = 18. Find side a."

Marcus's work:

  1. asinB=bsinA\frac{a}{\sin B} = \frac{b}{\sin A}
  2. a=18×sin65°sin50°a = \frac{18 \times \sin 65\degree}{\sin 50\degree}
  3. a=18×0.90630.7660=21.3a = \frac{18 \times 0.9063}{0.7660} = 21.3

What error did Marcus make, and what is the correct value of side a?

2.

Taylor solved this problem:

"In triangle ABC, angle A = 25°, side a = 6, and side b = 11. Find angle B."

Taylor's work:

  1. sinB=11×sin25°6=11×0.42266=0.7748\sin B = \frac{11 \times \sin 25\degree}{6} = \frac{11 \times 0.4226}{6} = 0.7748
  2. B=arcsin(0.7748)=50.8°B = \arcsin(0.7748) = 50.8\degree
  3. C=180°25°50.8°=104.2°C = 180\degree - 25\degree - 50.8\degree = 104.2\degree
  4. "There is one triangle."

What error did Taylor make?

F

Challenge / Extension

1.

In triangle ABC, angle A = 40°, side a=5a = 5, and side b=12b = 12. How many triangles can be formed?

2.

In triangle ABC, angle A = 42°, angle B = 73°, and side c=20c = 20. Find the area of the triangle. Round to the nearest tenth.

Hint: The area of a triangle is 12absinC\frac{1}{2}ab\sin C.

0 of 20 answered