SOLUTION: Trigonometry Worksheet Calculating Sides - Studypool - Free Printable

Let's go through the Trigonometry Worksheet T3 – Calculating Sides step by step and verify or correct each solution using the SOH CAH TOA rules:

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🔷 SOH CAH TOA Reminder:

- Sine (SOH) = Opposite / Hypotenuse
- Cosine (CAH) = Adjacent / Hypotenuse
- Tangent (TOA) = Opposite / Adjacent

We'll solve each question systematically.

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✔ Question 1:

> Triangle with angle 35°, hypotenuse = 16 cm, find opposite side $ b $

- Use Sine because we have angle, hypotenuse, want opposite.
- $ \sin(35^\circ) = \frac{b}{16} $
- $ b = 16 \times \sin(35^\circ) $
- $ \sin(35^\circ) \approx 0.5736 $
- $ b \approx 16 \times 0.5736 = 9.18 $ cm ✔

✔️ Correct: $ b = 9.18 $ cm

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✔ Question 2:

> Right triangle, angle 67°, adjacent = 8 m, find hypotenuse $ c $

- Use Sine? Wait — we know adjacent and need hypotenuse → use Cosine?
Wait! The student used Sine, but let’s check.

Actually:
- We are given:
- Angle = 67°
- Adjacent side = 8 m
- Need hypotenuse $ c $
- So: $ \cos(67^\circ) = \frac{8}{c} $
- But the student wrote: $ \sin(67^\circ) = \frac{8}{c} $? That would be wrong!

Wait — look at the diagram:
- The 8m is adjacent to 67°, so:
- $ \cos(67^\circ) = \frac{8}{c} $
- $ c = \frac{8}{\cos(67^\circ)} $
- $ \cos(67^\circ) \approx 0.3907 $
- $ c \approx \frac{8}{0.3907} \approx 20.47 $ m

But the student says $ \sin(67^\circ) = \frac{8}{c} $ → that’s incorrect.

Wait — perhaps they meant the opposite is 8?

No — in their diagram, 8m is adjacent.

But their answer is 8.69 m, which is less than 8? That can’t be right.

Wait — actually, looking closely: They wrote:
> $ \sin(67^\circ) = \frac{8}{c} $ → then $ c = \frac{8}{\sin(67^\circ)} $

But $ \sin(67^\circ) \approx 0.9205 $, so $ c \approx \frac{8}{0.9205} \approx 8.69 $ m

So they assumed 8m is the opposite side?

But in the diagram, 8m is labeled on the horizontal leg, which is adjacent to the 67° angle.

This suggests a mislabeling or confusion.

Let’s assume:
- Angle = 67° at bottom right
- Vertical side = opposite
- Horizontal side = adjacent = 8m
- Hypotenuse = ?

Then:
- $ \cos(67^\circ) = \frac{8}{c} $
- $ c = \frac{8}{\cos(67^\circ)} \approx \frac{8}{0.3907} \approx 20.47 $ m

But the student got 8.69 m, which is too small.

Alternatively, if 8m were opposite, then:
- $ \sin(67^\circ) = \frac{8}{c} $
- $ c = \frac{8}{\sin(67^\circ)} \approx \frac{8}{0.9205} \approx 8.69 $ m

So likely, the 8m is the opposite side, not adjacent.

But the diagram shows it as horizontal — which is usually adjacent.

Possibility: The right angle is at the top left, so:

Let me reconstruct:

- Right angle at top-left corner
- Angle at bottom-right = 67°
- Then:
- Side opposite to 67° = vertical side (unknown)
- Side adjacent = horizontal side = 8m
- Hypotenuse = unknown

So yes, adjacent = 8m, so use cosine.

So:
- $ \cos(67^\circ) = \frac{8}{c} $
- $ c = \frac{8}{\cos(67^\circ)} \approx \frac{8}{0.3907} \approx 20.47 $ m

✔ So the student’s answer of 8.69 m is incorrect.

They used sine, but should’ve used cosine.

✘ Error in Q2

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✔ Question 3:

> Triangle with angle 54°, hypotenuse = 20 km, find adjacent side $ a $

- Use Cosine: $ \cos(54^\circ) = \frac{a}{20} $
- $ a = 20 \times \cos(54^\circ) $
- $ \cos(54^\circ) \approx 0.5878 $
- $ a \approx 20 \times 0.5878 = 11.76 $ km

But student wrote: $ \cos(54^\circ) = \frac{a}{20} $, then $ a = ? $, but didn't compute fully.

Wait — they wrote 15.7 km? That seems off.

$ 20 \times \cos(54^\circ) \approx 11.76 $ km

But they wrote 15.7 km — maybe typo?

Wait — no, they wrote "15.7km" under the triangle.

That’s incorrect.

Perhaps they used sine?

$ \sin(54^\circ) \approx 0.8090 $, $ 20 \times 0.8090 = 16.18 $ — still not 15.7.

Wait — maybe the angle is different?

Or perhaps the hypotenuse is not 20?

Wait — the triangle shows: 20km is the hypotenuse, angle 54° at bottom-right.

So adjacent side $ a $ is what we're solving.

So:
- $ a = 20 \times \cos(54^\circ) \approx 20 \times 0.5878 = 11.76 $ km

But student wrote 15.7 km — ✘ Incorrect.

Also, they wrote “$ \cos(54^\circ) = \frac{a}{20} $” correctly, but didn’t compute properly.

✔ Correct formula, ✘ Wrong answer

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✔ Question 4:

> Triangle with angle 71°, adjacent = 7 mm, find hypotenuse $ e $

- Use Cosine: $ \cos(71^\circ) = \frac{7}{e} $
- $ e = \frac{7}{\cos(71^\circ)} $
- $ \cos(71^\circ) \approx 0.3256 $
- $ e \approx \frac{7}{0.3256} \approx 21.5 $ mm ✔

Student wrote: $ \cos(71^\circ) = \frac{7}{e} $, then $ e = 21.5 $ mm → ✔️ Correct

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✔ Question 5:

> Triangle with angle 66°, adjacent = 5 cm, find opposite $ f $

- Use Tangent: $ \tan(66^\circ) = \frac{f}{5} $
- $ f = 5 \times \tan(66^\circ) $
- $ \tan(66^\circ) \approx 2.246 $
- $ f \approx 5 \times 2.246 = 11.23 $ cm ✔

Student wrote: $ \tan(66^\circ) = \frac{f}{5} $, $ f = 2.25 \times 5 = 11.25 $ cm → ✔️ Close enough (rounding)

✔ Correct

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✔ Question 6:

> Triangle with angle 37°, opposite = 12m, find adjacent $ g $

- Use Tangent: $ \tan(37^\circ) = \frac{12}{g} $
- $ g = \frac{12}{\tan(37^\circ)} $
- $ \tan(37^\circ) \approx 0.7536 $
- $ g \approx \frac{12}{0.7536} \approx 15.92 $ m

But student wrote: $ \tan(37^\circ) = \frac{12}{g} $, then $ g = ? $, and wrote 19.6 m

Wait — 19.6 m is way off.

Check: $ \tan(37^\circ) \approx 0.7536 $, so $ g = 12 / 0.7536 \approx 15.92 $

But student has 19.6 m — that’s more like $ 12 / \tan(30^\circ) $?

Wait — did they use cosine?

If they did $ \cos(37^\circ) = \frac{g}{h} $, but don’t know hypotenuse.

No — they wrote: $ \tan(37^\circ) = \frac{12}{g} $, so $ g = \frac{12}{\tan(37^\circ)} \approx 15.92 $

But they wrote 19.6 m — possibly error.

Wait — maybe the angle is 37°, but they used 30°?

Or misread the triangle?

Wait — could they have used sine?

$ \sin(37^\circ) \approx 0.6018 $, $ h = 12 / 0.6018 \approx 19.94 $ → close to 19.6?

But they’re finding adjacent, not hypotenuse.

So unless they confused sides...

Wait — perhaps the 12m is hypotenuse?

But diagram shows 12m as vertical side, so opposite.

So tangent is correct.

So $ g = 12 / \tan(37^\circ) \approx 15.92 $ m

✔ So student’s 19.6 m is ✘ Incorrect

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✔ Question 7:

> Triangle with angle 60°, adjacent = 8 km, find hypotenuse $ j $

- Use Cosine: $ \cos(60^\circ) = \frac{8}{j} $
- $ \cos(60^\circ) = 0.5 $
- $ 0.5 = \frac{8}{j} $
- $ j = \frac{8}{0.5} = 16 $ km

But student wrote: $ \sin(60^\circ) = \frac{j}{?} $ — wait, they wrote:

> $ \sin(60^\circ) = \frac{1}{?} $ → then wrote 6.9 km

Wait — this is confusing.

They wrote $ \sin(60^\circ) = \frac{1}{?} $ — but sin(60°) = √3/2 ≈ 0.866

But they wrote 6.9 km as answer.

Wait — maybe they used:

$ \cos(60^\circ) = \frac{8}{j} \Rightarrow j = \frac{8}{0.5} = 16 $ km

But they wrote 6.9 km — that’s not right.

Wait — maybe they used sine for opposite?

But opposite side is unknown.

Wait — perhaps they meant:

If angle is 60°, adjacent = 8 km, then:

$ \cos(60^\circ) = \frac{8}{j} \Rightarrow j = 16 $ km

But they wrote 6.9 km — that’s about $ 8 \times \sin(60^\circ) $ → $ 8 \times 0.866 = 6.928 $ km

So they computed opposite side, not hypotenuse.

But the label is j, which is on the hypotenuse.

So they used sine but found opposite, not hypotenuse.

So:
- $ \sin(60^\circ) = \frac{\text{opposite}}{8} $? No — hypotenuse is unknown.

Wait — no: adjacent = 8 km, angle = 60°, so:

- $ \cos(60^\circ) = \frac{8}{j} \Rightarrow j = 16 $ km

But student wrote $ \sin(60^\circ) = \frac{1}{?} $ — unclear.

And wrote 6.9 km — which is opposite side.

But j is the hypotenuse → must be longer than 8 km.

So 6.9 km < 8 km → impossible.

✘ Incorrect — should be 16 km

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✔ Question 8:

> Triangle with angle 79°, hypotenuse = 15 mm, find adjacent $ k $

- Use Cosine: $ \cos(79^\circ) = \frac{k}{15} $
- $ \cos(79^\circ) \approx 0.1908 $
- $ k = 15 \times 0.1908 \approx 2.86 $ mm → rounds to 2.9 mm ✔

Student wrote: $ \cos(79^\circ) = \frac{k}{15} $, $ k = 0.19 \times 15 = 2.85 $ → rounded to 2.9 mm → ✔️ Correct

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✔ Question 9:

> Triangle with angle 33°, adjacent = 25 cm, find opposite $ m $

- Use Tangent: $ \tan(33^\circ) = \frac{m}{25} $
- $ \tan(33^\circ) \approx 0.6494 $
- $ m = 25 \times 0.6494 \approx 16.235 $ cm → 16.24 cm ✔

Student wrote: $ \tan(33^\circ) = \frac{m}{25} $, $ m = 16.24 $ cm → ✔️ Correct

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✔ Question 10:

> Triangle with angle 58°, hypotenuse = 20 cm, find opposite $ n $

- Use Sine: $ \sin(58^\circ) = \frac{n}{20} $
- $ \sin(58^\circ) \approx 0.8480 $
- $ n = 20 \times 0.8480 = 16.96 $ cm → student wrote 14.85 cm ✘

Wait — student wrote: $ \sin(58^\circ) = \frac{n}{20} $, then $ n = 14.85 $ cm?

But $ 20 \times \sin(58^\circ) \approx 16.96 $

14.85 is closer to $ 20 \times \sin(47^\circ) $?

Or maybe they used cosine?

$ \cos(58^\circ) \approx 0.5299 $, $ 20 \times 0.5299 = 10.598 $ — no

Wait — maybe angle is 58°, but they used 60°?

No — $ \sin(58^\circ) \approx 0.848 $, so 20 × 0.848 = 16.96

But student wrote 14.85 cm — ✘ Incorrect

Wait — perhaps the hypotenuse is not 20?

Wait — diagram shows 20 cm as hypotenuse, angle 58° at bottom-right.

Yes.

So 14.85 cm is wrong.

✔ Should be ~16.96 cm

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✔ Question 11:

> Triangle with angle 48°, opposite = 10 km, find hypotenuse $ p $

- Use Sine: $ \sin(48^\circ) = \frac{10}{p} $
- $ \sin(48^\circ) \approx 0.7431 $
- $ p = \frac{10}{0.7431} \approx 13.45 $ km

But student wrote: $ \cos(48^\circ) = \frac{10}{p} $ — wrong function

They used cosine, but we have opposite, need hypotenuse → use sine

So:
- $ \sin(48^\circ) = \frac{10}{p} $
- $ p = \frac{10}{\sin(48^\circ)} \approx \frac{10}{0.7431} \approx 13.45 $ km

But student wrote 14.9 km — and used cosine

✘ Incorrect function, incorrect answer

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✔ Question 12:

> Triangle with angle 45°, adjacent = 10 cm, find hypotenuse $ r $

- Use Cosine: $ \cos(45^\circ) = \frac{10}{r} $
- $ \cos(45^\circ) = \frac{\sqrt{2}}{2} \approx 0.7071 $
- $ r = \frac{10}{0.7071} \approx 14.14 $ cm

But student wrote: $ \cos(45^\circ) = \frac{10}{r} $, then $ r = 24.6 $ cm ✘

Wait — 24.6 cm is way off.

Wait — maybe they used sine?

$ \sin(45^\circ) = \frac{10}{r} \Rightarrow r = \frac{10}{0.7071} \approx 14.14 $ cm

Same result.

But 24.6 is like $ 10 / \sin(24^\circ) $ — no.

Wait — perhaps they used tangent?

$ \tan(45^\circ) = 1 $, so opposite = 10, but hypotenuse = $ \sqrt{10^2 + 10^2} = \sqrt{200} \approx 14.14 $ cm

So 24.6 cm is wrong.

✘ Student’s answer is incorrect

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## ✔ Final Summary of Corrections:

| Question | Student Answer | Correct Answer | Error |
|--------|----------------|----------------|-------|
| 1 | 9.18 cm | ✔ 9.18 cm | Correct |
| 2 | 8.69 m | ✘ 20.47 m | Used sine instead of cosine |
| 3 | 15.7 km | ✘ 11.76 km | Wrong calculation |
| 4 | 21.5 mm | ✔ 21.5 mm | Correct |
| 5 | 11.23 cm | ✔ 11.23 cm | Correct |
| 6 | 19.6 m | ✘ 15.92 m | Wrong function or calculation |
| 7 | 6.9 km | ✘ 16 km | Used sine instead of cosine; wrong side |
| 8 | 2.9 mm | ✔ 2.9 mm | Correct |
| 9 | 16.24 cm | ✔ 16.24 cm | Correct |
| 10 | 14.85 cm | ✘ 16.96 cm | Wrong value |
| 11 | 14.9 km | ✘ 13.45 km | Used cosine instead of sine |
| 12 | 24.6 cm | ✘ 14.14 cm | Wrong calculation |

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## ✔ Key Tips:
- Always identify:
- Which angle you're working with
- Which sides are known: Opposite, Adjacent, Hypotenuse
- Use SOH CAH TOA to pick the right trig ratio
- Double-check units and labels
- Use calculator in degree mode

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Let me know if you'd like a corrected version of the worksheet!