Missing Sides worksheet for finding unknown side lengths in triangles using given angles and side measurements.

Let’s solve each problem one by one. We’ll use the Law of Sines for all these triangles because we’re given two angles and one side (or can find the third angle), and we need to find a missing side.

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 the sides
- $A$, $B$, $C$ are the angles opposite those sides

We’ll round answers to the nearest tenth as instructed.

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Problem 1: Triangle PQR

Given:
- Angle at Q = 49°
- Angle at R = 75°
- Side QR = 13 (this is side opposite angle P)
- Need to find PQ (side opposite angle R)

First, find angle at P:

Angle P = 180° - 49° - 75° = 56°

Now apply Law of Sines:

$\frac{PQ}{\sin 75^\circ} = \frac{QR}{\sin 56^\circ}$

→ $\frac{PQ}{\sin 75^\circ} = \frac{13}{\sin 56^\circ}$

Calculate sin values:

- sin 75° ≈ 0.9659
- sin 56° ≈ 0.8290

So:

PQ = $13 × \frac{0.9659}{0.8290}$ ≈ 13 × 1.1651 ≈ 15.1

✔ PQ ≈ 15.1

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Problem 2: Triangle ABC

Given:
- Angle at A = 63°
- Angle at B = 22°
- Side BC = 24 (opposite angle A)
- Need to find AC (opposite angle B)

Find angle at C:

Angle C = 180° - 63° - 22° = 95°

Law of Sines:

$\frac{AC}{\sin 22^\circ} = \frac{BC}{\sin 63^\circ}$

→ $\frac{AC}{\sin 22^\circ} = \frac{24}{\sin 63^\circ}$

Sin values:

- sin 22° ≈ 0.3746
- sin 63° ≈ 0.8910

AC = $24 × \frac{0.3746}{0.8910}$ ≈ 24 × 0.4204 ≈ 10.1

✔ AC ≈ 10.1

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Problem 3: Triangle XYZ

Given:
- Angle at X = 73.6°
- Angle at Z = 63°
- Side XZ = 73.9 (opposite angle Y)
- Need to find YZ (opposite angle X)

Find angle at Y:

Angle Y = 180° - 73.6° - 63° = 43.4°

Law of Sines:

$\frac{YZ}{\sin 73.6^\circ} = \frac{XZ}{\sin 43.4^\circ}$

→ $\frac{YZ}{\sin 73.6^\circ} = \frac{73.9}{\sin 43.4^\circ}$

Sin values:

- sin 73.6° ≈ 0.9595
- sin 43.4° ≈ 0.6870

YZ = $73.9 × \frac{0.9595}{0.6870}$ ≈ 73.9 × 1.3967 ≈ 103.2

✔ YZ ≈ 103.2

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Problem 4: Triangle VWU

Given:
- Angle at U = 28°
- Angle at W = ? → Wait, let’s check diagram.

Actually, from diagram:

Triangle VWU:

- Side VU = 34 (opposite angle W)
- Side WU = 12 (opposite angle V)
- Angle at U = 28°
- Need to find VW (side opposite angle U)

Wait — actually, looking again:

Points: V, W, U

Side VU = 34
Side WU = 12
Angle at U = 28°
Need to find VW

This is SAS? No — we have two sides and included angle? Actually, no — we have two sides and non-included angle? Let me re-read.

Actually, in triangle VWU:

We know:
- Side VU = 34
- Side WU = 12
- Angle at U = 28°

But angle at U is between sides VU and WU? Yes! So this is SAS — but we don’t have Law of Cosines here? The worksheet is titled “Missing Sides” and likely expects Law of Sines — but with SAS, we need Law of Cosines first to find the third side, then maybe Law of Sines for others.

Wait — perhaps I misread the diagram.

Looking back: In problem 4, it shows triangle VWU with:

- Side VU = 34
- Side WU = 12
- Angle at U = 28°
- And we need to find VW

Yes — so angle at U is between sides VU and WU → that’s SAS.

So we must use Law of Cosines to find side VW.

Law of Cosines:

$c^2 = a^2 + b^2 - 2ab \cos C$

Here, side VW is opposite angle U, so:

VW² = VU² + WU² - 2·VU·WU·cos(angle U)

→ VW² = 34² + 12² - 2·34·12·cos(28°)

Calculate:

34² = 1156
12² = 144
Sum = 1300

2·34·12 = 816
cos(28°) ≈ 0.8829

So:

VW² = 1300 - 816 × 0.8829
= 1300 - 720.4464
≈ 579.5536

VW ≈ √579.5536 ≈ 24.1

✔ VW ≈ 24.1

*(Note: Even though other problems used Law of Sines, this one requires Law of Cosines because we have SAS.)*

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Problem 5: Triangle EFD

Given:
- Angle at F = 96°
- Angle at D = 20°
- Side FD = 84 (opposite angle E)
- Need to find EF (opposite angle D)

Find angle at E:

Angle E = 180° - 96° - 20° = 64°

Law of Sines:

$\frac{EF}{\sin 20^\circ} = \frac{FD}{\sin 64^\circ}$

→ $\frac{EF}{\sin 20^\circ} = \frac{84}{\sin 64^\circ}$

Sin values:

- sin 20° ≈ 0.3420
- sin 64° ≈ 0.8988

EF = $84 × \frac{0.3420}{0.8988}$ ≈ 84 × 0.3805 ≈ 32.0

✔ EF ≈ 32.0

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Problem 6: Triangle LMN

Given:
- Angle at M = 46°
- Angle at N = 72°
- Side LN = 84 (opposite angle M)
- Need to find MN (opposite angle L)

Find angle at L:

Angle L = 180° - 46° - 72° = 62°

Law of Sines:

$\frac{MN}{\sin 62^\circ} = \frac{LN}{\sin 46^\circ}$

→ $\frac{MN}{\sin 62^\circ} = \frac{84}{\sin 46^\circ}$

Sin values:

- sin 62° ≈ 0.8829
- sin 46° ≈ 0.7193

MN = $84 × \frac{0.8829}{0.7193}$ ≈ 84 × 1.2275 ≈ 103.1

✔ MN ≈ 103.1

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Final Answer:

1) PQ = 15.1
2) AC = 10.1
3) YZ = 103.2
4) VW = 24.1
5) EF = 32.0
6) MN = 103.1