← Unit 4 Solving Right-Angled Triangles

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MA5-TRG-C-01

Solving Right-Angled Triangles

⏱ 25 min📚 Year 9📈
Think First

A right triangle has one angle $35°$ and hypotenuse 8 cm. How would you find the opposite side?

💡 Revision: Ensure you know SOH CAH TOA and can use a calculator in degree mode.

Learning Intentions

Know

  • Selecting the correct trig ratio
  • Rearranging equations
  • Finding angles with inverse trig

Understand

  • Why the choice of ratio depends on what is given and what is sought
  • How to check answers using a different ratio

Can Do

  • Find an unknown side given an angle and one side
  • Find an unknown angle given two sides
  • Verify answers by substitution
Inverse trigRearrangeSubstituteVerify
Learn Phase
1

Finding an Unknown Side

Choose the right ratio

To find an unknown side:

  1. Label the sides relative to the known angle
  2. Identify which sides you know and which you want
  3. Choose the ratio that connects them
  4. Set up the equation and solve

Example: Find $x$ given angle $40°$, hypotenuse 10, and $x$ is adjacent.

$cos 40° = x/10$ → $x = 10 imes cos 40° approx 7.66$

2

Finding an Unknown Angle

Use inverse trig functions

To find an unknown angle when two sides are known:

  1. Label the sides relative to the unknown angle
  2. Choose the ratio connecting the known sides
  3. Use the inverse function

Example: Opposite = 5, hypotenuse = 10. Find the angle.

$sin heta = 5/10 = 0.5$

$ heta = sin^{-1}(0.5) = 30°$

3

Checking Your Answer

Verify with another ratio

Always check your work:

  • Does the angle sum to 180°?
  • Does Pythagoras hold?
  • Can you recalculate using a different ratio?

Example: If you found $x = 7.66$ using cosine, check with sine:

$sin 40° = ext{opp}/10$ → opp $= 10 imes sin 40° approx 6.43$

Verify: $7.66^2 + 6.43^2 approx 58.7 + 41.3 = 100 = 10^2$ ✓

✓

Check Understanding

Try it yourself

A right triangle has angle $50°$ and adjacent side 6 cm. Find the opposite side.

Worked Example

Solving Right-Angled Triangles

1

Find $x$ given angle $30°$, hypotenuse 12, $x$ opposite.

$sin 30° = x/12$

$x = 12 imes 0.5 = 6$

2

Find angle $ heta$ given opposite = 8, adjacent = 6.

$ an heta = 8/6 = 4/3$

$ heta = an^{-1}(4/3) approx 53.1°$

3

A ladder 5 m long leans against a wall at $60°$ to the ground. How high does it reach?

$sin 60° = h/5$

$h = 5 imes sqrt{3}/2 approx 4.33$ m

Common Misconceptions

✕

Use sine when you know adjacent and want hypotenuse. No — sine connects opposite and hypotenuse. Use cosine for adjacent and hypotenuse.

✕

$sin^{-1}(x) = 1/sin(x)$. No — $sin^{-1}$ is the inverse sine (arcsin), not the reciprocal. The reciprocal is $csc x$.

✕

Round intermediate values before the final answer. No — keep full precision throughout and round only the final answer.

Your Turn

Practice — Solving Triangles

Work through each question in your book or digitally. Answers are in the Questions phase.

1Find $x$ if angle $= 25°$, hypotenuse $= 10$, $x$ adjacent.
2Find angle $ heta$ if opposite $= 7$, hypotenuse $= 14$.
3A ramp rises 2 m over 10 m horizontal. Find the angle of inclination.
Real-World Anchor

Construction and Carpentry

Builders use trigonometry to calculate roof pitches, stair angles, and diagonal bracing. A roof pitched at 30° will have specific height-to-span ratios that carpenters calculate using tangent.

📓 Copy Into Your Books

▼

Finding a side

  • Label sides
  • Choose ratio (SOH CAH TOA)
  • Set up equation
  • Solve

Finding an angle

  • Label sides
  • Set up ratio
  • Use inverse function ($sin^{-1}$ etc.)

Check

  • Pythagoras check
  • Angle sum = 180°
  • Recalculate with different ratio
Questions Phase
Check Your Understanding
Answer all questions correctly to unlock the Game phase.
To find the opposite side given hypotenuse and angle, use:
$sin^{-1}(0.5) = $
A triangle has opposite = 3, hypotenuse = 5. The angle is:
$sin^{-1}(x)$ means:
To find adjacent given opposite and angle, use:
A 10 m ladder at $70°$ to ground reaches:
Given adjacent = 8, hypotenuse = 10, find angle:
$ an^{-1}(1) = $
1Find $x$ given angle $= 35°$, adjacent $= 8$ cm, $x$ = opposite.
2A kite string is 50 m long at $40°$ to horizontal. How high is the kite?
3Explain why you should not round intermediate values in trig calculations.

Comprehensive Answers

1Find $x$ with $35°$, adjacent $= 8$, $x$ opposite.
$x = 8 imes an 35° approx 5.60$ cm.
2Kite 50 m at $40°$. Height?
$h = 50 imes sin 40° approx 32.1$ m.
3Why not round intermediates?
Rounding introduces error that compounds. Keep full precision until final answer.
MC 1Find opposite given hyp and angle.
Sine. Answer: B
MC 2$sin^{-1}(0.5)$.
$30°$. Answer: B
MC 3Opposite 3, hyp 5.
$sin^{-1}(3/5) approx 36.9°$. Answer: B
MC 4$sin^{-1}(x)$ meaning.
Angle whose sine is $x$. Answer: B
MC 5Find adjacent given opp and angle.
Tangent: $ ext{adj} = ext{opp}/ an heta$. Answer: C
MC 610 m ladder at $70°$.
$10 imes sin 70° approx 9.4$ m. Answer: A
MC 7Adjacent 8, hyp 10.
$cos^{-1}(8/10) = cos^{-1}(0.8) approx 36.9°$. Answer: A
MC 8$ an^{-1}(1)$.
$45°$. Answer: B
SA 1Find $x$ with $35°$, adj 8.
$x = 8 imes an 35° approx 5.60$ cm.
SA 2Kite height.
$50 imes sin 40° approx 32.1$ m.
SA 3Rounding intermediates.
Compounding error. Keep full precision.
Game Phase
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