Trigonometry - Missing Angles worksheet with right-angled triangles and side lengths for calculating unknown angles.

Let’s solve each problem step by step. We’re finding missing angles in right-angled triangles using trigonometry (sine, cosine, tangent). Remember:

- SOH CAH TOA:
- Sin = Opposite / Hypotenuse
- Cos = Adjacent / Hypotenuse
- Tan = Opposite / Adjacent

We’ll use inverse functions (sin⁻¹, cos⁻¹, tan⁻¹) on our calculator to find the angle.

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Problem 1: Find angles a, b, c, d, e, f

#### Triangle with angle a:
Right triangle. Sides: opposite = 18 cm, hypotenuse = 24 cm → use sin

> sin(a) = opposite/hypotenuse = 18/24 = 0.75
> a = sin⁻¹(0.75) ≈ 48.6°

✔ Check: 18² + ?² = 24² → adjacent = √(576 - 324) = √252 ≈ 15.87 — not needed since we have opp and hyp.

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#### Triangle with angle b:
Sides: adjacent = 12 cm, opposite = 15 cm → use tan

> tan(b) = opposite/adjacent = 15/12 = 1.25
> b = tan⁻¹(1.25) ≈ 51.3°

Wait — let’s double-check labeling. Angle b is at top left. The side next to it (adjacent) is 12 cm (vertical), and opposite is 15 cm (horizontal base). Yes, correct.

But wait — actually, looking again: if angle b is at the top vertex, then the side *opposite* to it is the bottom side (15 cm), and the side *adjacent* is the vertical leg (12 cm). So yes, tan(b) = 15/12 → correct.

✔ b ≈ 51.3°

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#### Triangle with angle c:
This one has sides: adjacent = 6 cm, hypotenuse = 11 cm → use cos

Angle c is at bottom right corner. The side next to it (adjacent) is 6 cm (vertical leg), hypotenuse is 11 cm.

> cos(c) = adjacent/hypotenuse = 6/11 ≈ 0.5455
> c = cos⁻¹(0.5455) ≈ 56.9°

✔ Correct.

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#### Triangle with angle d:
Vertical side = 41 cm (adjacent to angle d), horizontal top = 12 cm (opposite to angle d)

So for angle d (at bottom):

> tan(d) = opposite/adjacent = 12/41 ≈ 0.2927
> d = tan⁻¹(0.2927) ≈ 16.3°

✔ Correct.

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#### Triangle with angle e:
Hypotenuse = 20 cm, opposite side to angle e = 16 cm? Wait — let's see.

Angle e is at top left. The side opposite to it is the vertical leg = 16 cm. Hypotenuse = 20 cm.

So:

> sin(e) = opposite/hypotenuse = 16/20 = 0.8
> e = sin⁻¹(0.8) ≈ 53.1°

✔ Correct.

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#### Triangle with angle f:
Right angle at top. Side adjacent to f = 42 cm, hypotenuse = 170 cm? Wait — no.

Actually, angle f is at bottom right. The side adjacent to it is 42 cm (top leg), and hypotenuse is 170 cm (longest side).

Wait — but 42 and 170 — that would make the other leg huge. Let me check:

If right angle is at top, then legs are 42 cm and ???, hypotenuse 170 cm.

Then for angle f (bottom right):

Adjacent side to f = 42 cm? No — actually, from angle f, the adjacent side is the bottom leg (unknown), and opposite is 42 cm? That doesn’t match.

Wait — better approach: label properly.

In the diagram: right angle at top. One leg = 42 cm (going down-right), hypotenuse = 170 cm (from bottom-left to top-right). Angle f is at bottom-right vertex.

So from angle f:

- Opposite side = 42 cm (the leg going up to right angle)
- Hypotenuse = 170 cm

→ So sin(f) = opposite/hypotenuse = 42/170 ≈ 0.2471
→ f = sin⁻¹(0.2471) ≈ 14.3°

Alternatively, maybe adjacent? Let’s think.

If angle f is at bottom right, and right angle is at top, then:

- Side between angle f and right angle = 42 cm → this is adjacent to angle f?
No — actually, the side connecting angle f to the right angle is one leg — which is adjacent to angle f only if it’s next to it along the angle.

Actually, standard: in right triangle, for any acute angle:

- Opposite = side across from angle
- Adjacent = side next to angle (not hypotenuse)
- Hypotenuse = longest side

So for angle f (bottom right):

- Opposite = the leg going up to the right angle = 42 cm
- Adjacent = the bottom leg (unknown)
- Hypotenuse = 170 cm

So yes — sin(f) = 42/170 → f ≈ 14.3°

But wait — 42² + x² = 170² → x² = 28900 - 1764 = 27136 → x ≈ 164.7 cm — so adjacent is ~164.7 cm.

Then tan(f) = 42 / 164.7 ≈ 0.255 → f ≈ 14.3° — same answer.

Or cos(f) = adjacent/hypotenuse = 164.7/170 ≈ 0.9688 → f ≈ 14.3° — consistent.

✔ So f ≈ 14.3°

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Problem 2: Find angles x and y

#### Triangle with angle x:
Isosceles triangle (two equal sides marked), height = 35 cm, base = 26 cm.

The height splits the base into two equal parts: 13 cm each.

So now we have a right triangle with:

- Opposite to half of angle x? Wait — angle x is at the apex.

Actually, the height creates two right triangles. In each, angle at apex is half of x.

So in one right triangle:

- Opposite = 13 cm (half-base)
- Adjacent = 35 cm (height)

So tan(x/2) = opposite/adjacent = 13/35 ≈ 0.3714
→ x/2 = tan⁻¹(0.3714) ≈ 20.4°
→ x ≈ 40.8°

✔ Correct.

Alternative: could use sin or cos, but tan is easiest here.

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#### Rectangle with diagonal and angle y:
Rectangle: width = 90 m, diagonal = 110 m. Angle y is at top right, between diagonal and top side.

So in the right triangle formed:

- Adjacent to angle y = 90 m (top side)
- Hypotenuse = 110 m (diagonal)

→ cos(y) = adjacent/hypotenuse = 90/110 ≈ 0.8182
→ y = cos⁻¹(0.8182) ≈ 35.1°

Check: opposite side = √(110² - 90²) = √(12100 - 8100) = √4000 ≈ 63.25 m
Then sin(y) = 63.25/110 ≈ 0.575 → y ≈ 35.1° — matches.

✔ Correct.

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## Final Answers:

Problem 1:

a = 48.6°
b = 51.3°
c = 56.9°
d = 16.3°
e = 53.1°
f = 14.3°

Problem 2:

x = 40.8°
y = 35.1°

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Final Answer:
a = 48.6°, b = 51.3°, c = 56.9°, d = 16.3°, e = 53.1°, f = 14.3°, x = 40.8°, y = 35.1°