Law of Sines Practice Worksheet with eight problems involving triangles, angles, and sides.

Let’s solve each problem one by one using the Law of Sines.

The Law of Sines says:

> In any triangle,
> $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

Where:
- $a$, $b$, $c$ are side lengths
- $A$, $B$, $C$ are the angles opposite those sides

We’ll use this to find missing sides or angles. Remember: in a triangle, all three angles add up to 180° — so if two angles are given, we can always find the third.

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Problem 1:

Triangle with angles 53° and 40°, side opposite 40° is 5, find side x opposite 53°.

First, find the third angle:
→ 180° - 53° - 40° = 87°

Now apply Law of Sines:

$\frac{x}{\sin 53^\circ} = \frac{5}{\sin 40^\circ}$

Solve for x:

$x = \frac{5 \cdot \sin 53^\circ}{\sin 40^\circ}$

Use calculator (make sure it’s in degree mode):

sin 53° ≈ 0.7986
sin 40° ≈ 0.6428

→ $x ≈ \frac{5 \cdot 0.7986}{0.6428} ≈ \frac{3.993}{0.6428} ≈ 6.21$

✔ Answer: x ≈ 6.21

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Problem 2:

Angles 128° and 32°, side opposite 32° is 10, find side x opposite 128°.

Wait — let’s check: sum of angles must be 180°.

Given angles: 128° + 32° = 160° → third angle = 20°

But wait — look at the diagram: side labeled “10” is opposite the 32° angle? Actually, looking again — the side labeled “10” is between the 128° and 32° angles? No — actually, in standard notation, side opposite an angle is across from it.

In problem 2: the side labeled “10” is opposite the angle that’s NOT labeled — which must be the third angle.

Wait — let’s re-read the diagram description.

Actually, in problem 2: the triangle has angles 128° and 32°, and side “10” is adjacent to both? That doesn’t make sense.

Looking carefully: the side labeled “10” is opposite the unlabeled angle. The side labeled “x” is opposite the 32° angle.

So first, find the third angle:

180° - 128° - 32° = 20°

So now:

Side opposite 20° is 10
Side opposite 32° is x

Apply Law of Sines:

$\frac{x}{\sin 32^\circ} = \frac{10}{\sin 20^\circ}$

→ $x = \frac{10 \cdot \sin 32^\circ}{\sin 20^\circ}$

sin 32° ≈ 0.5299
sin 20° ≈ 0.3420

→ $x ≈ \frac{10 \cdot 0.5299}{0.3420} ≈ \frac{5.299}{0.3420} ≈ 15.49$

✔ Answer: x ≈ 15.49

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Problem 3:

Angles 99° and 50°, side opposite 50° is 9, find side x opposite 99°.

Third angle: 180 - 99 - 50 = 31°

Law of Sines:

$\frac{x}{\sin 99^\circ} = \frac{9}{\sin 50^\circ}$

→ $x = \frac{9 \cdot \sin 99^\circ}{\sin 50^\circ}$

sin 99° ≈ sin(90+9) = cos 9° ≈ 0.9877
sin 50° ≈ 0.7660

→ $x ≈ \frac{9 \cdot 0.9877}{0.7660} ≈ \frac{8.8893}{0.7660} ≈ 11.60$

✔ Answer: x ≈ 11.60

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Problem 4:

Angles 62° and 65°, side opposite 65° is 6, find side x opposite 62°.

Third angle: 180 - 62 - 65 = 53°

Law of Sines:

$\frac{x}{\sin 62^\circ} = \frac{6}{\sin 65^\circ}$

→ $x = \frac{6 \cdot \sin 62^\circ}{\sin 65^\circ}$

sin 62° ≈ 0.8829
sin 65° ≈ 0.9063

→ $x ≈ \frac{6 \cdot 0.8829}{0.9063} ≈ \frac{5.2974}{0.9063} ≈ 5.84$

✔ Answer: x ≈ 5.84

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Problem 5:

Angles 110° and 46°, side opposite 46° is 15, find side x opposite 110°.

Third angle: 180 - 110 - 46 = 24°

Law of Sines:

$\frac{x}{\sin 110^\circ} = \frac{15}{\sin 46^\circ}$

→ $x = \frac{15 \cdot \sin 110^\circ}{\sin 46^\circ}$

sin 110° = sin(180-70) = sin 70° ≈ 0.9397
sin 46° ≈ 0.7193

→ $x ≈ \frac{15 \cdot 0.9397}{0.7193} ≈ \frac{14.0955}{0.7193} ≈ 19.59$

✔ Answer: x ≈ 19.59

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Problem 6:

Angles 81° and 61°, side opposite 81° is 7, find side x opposite 61°.

Third angle: 180 - 81 - 61 = 38°

Law of Sines:

$\frac{x}{\sin 61^\circ} = \frac{7}{\sin 81^\circ}$

→ $x = \frac{7 \cdot \sin 61^\circ}{\sin 81^\circ}$

sin 61° ≈ 0.8746
sin 81° ≈ 0.9877

→ $x ≈ \frac{7 \cdot 0.8746}{0.9877} ≈ \frac{6.1222}{0.9877} ≈ 6.20$

✔ Answer: x ≈ 6.20

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Problem 7: Find ALL missing sides and angles.

Given: one side = 12, angles 41° and 76°

First, find third angle: 180 - 41 - 76 = 63°

Label the triangle:

Let’s say:
- Angle A = 41°, opposite side a = ?
- Angle B = 76°, opposite side b = ?
- Angle C = 63°, opposite side c = 12

Wait — actually, the side labeled 12 is opposite which angle? Looking at diagram: side 12 is between 41° and 76°? So it’s opposite the 63° angle.

Yes — so side opposite 63° is 12.

Now find side opposite 41° (call it x):

$\frac{x}{\sin 41^\circ} = \frac{12}{\sin 63^\circ}$

→ $x = \frac{12 \cdot \sin 41^\circ}{\sin 63^\circ}$

sin 41° ≈ 0.6561
sin 63° ≈ 0.8910

→ $x ≈ \frac{12 \cdot 0.6561}{0.8910} ≈ \frac{7.8732}{0.8910} ≈ 8.84$

Now find side opposite 76° (call it y):

$\frac{y}{\sin 76^\circ} = \frac{12}{\sin 63^\circ}$

→ $y = \frac{12 \cdot \sin 76^\circ}{\sin 63^\circ}$

sin 76° ≈ 0.9703

→ $y ≈ \frac{12 \cdot 0.9703}{0.8910} ≈ \frac{11.6436}{0.8910} ≈ 13.07$

✔ Answers:
- Third angle: 63°
- Side opposite 41°: ≈ 8.84
- Side opposite 76°: ≈ 13.07

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Problem 8: Find ALL missing sides and angles.

Given: angles 95° and 48°, side opposite 48° is 16

First, third angle: 180 - 95 - 48 = 37°

Side opposite 48° is 16 → so we can find other sides.

Find side opposite 95° (call it x):

$\frac{x}{\sin 95^\circ} = \frac{16}{\sin 48^\circ}$

→ $x = \frac{16 \cdot \sin 95^\circ}{\sin 48^\circ}$

sin 95° ≈ sin(90+5) = cos 5° ≈ 0.9962
sin 48° ≈ 0.7431

→ $x ≈ \frac{16 \cdot 0.9962}{0.7431} ≈ \frac{15.9392}{0.7431} ≈ 21.45$

Find side opposite 37° (call it y):

$\frac{y}{\sin 37^\circ} = \frac{16}{\sin 48^\circ}$

→ $y = \frac{16 \cdot \sin 37^\circ}{\sin 48^\circ}$

sin 37° ≈ 0.6018

→ $y ≈ \frac{16 \cdot 0.6018}{0.7431} ≈ \frac{9.6288}{0.7431} ≈ 12.96$

✔ Answers:
- Third angle: 37°
- Side opposite 95°: ≈ 21.45
- Side opposite 37°: ≈ 12.96

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Final Answer:

1. x ≈ 6.21
2. x ≈ 15.49
3. x ≈ 11.60
4. x ≈ 5.84
5. x ≈ 19.59
6. x ≈ 6.20
7. Missing angle: 63°; missing sides: ≈8.84 and ≈13.07
8. Missing angle: 37°; missing sides: ≈21.45 and ≈12.96