N.B.Nota Bene

Home  /  Standard Maths  /  Trigonometry & Circles

Right-Angled Triangles

A right angle locks the sides and angles of a triangle together. With Pythagoras for the sides and SOHCAHTOA for the ratios, two known pieces are enough to unlock a missing side or a missing angle.

MYP 5Standard MathsTrigonometryCriteria A · C · D~10 min read

Written by Fayek Jawad Haider · MYP student

Pythagoras' theorem

Before the ratios, start with the sides. Pythagoras' theorem links the three sides of any right-angled triangle: the square of the hypotenuse equals the sum of the squares of the other two sides.

Hypotenuse
The longest side, always opposite the right angle. Pythagoras only works with a right angle present.
\[ a^2 + b^2 = c^2 \]

Here \(c\) is the hypotenuse and \(a\) and \(b\) are the two shorter sides. Use it when you know two sides and want the third, and no angle other than the right angle is involved.

Worked example

A right-angled triangle has shorter sides of 5 cm and 12 cm. Find the hypotenuse.

1
The hypotenuse is the unknown, so use \(c^2 = a^2 + b^2\).
2
Substitute: \(c^2 = 5^2 + 12^2 = 25 + 144 = 169\).
3
Square root both sides: \(c = \sqrt{169} = 13\).
Hypotenuse \(= 13\) cm

Labelling and SOHCAHTOA

The trig ratios only work once the sides are named relative to the angle you are using. Pick the angle in question (never the right angle) and label from there: the opposite is across from it, the adjacent is next to it, and the hypotenuse is opposite the right angle.

\[ \sin\theta = \frac{\text{O}}{\text{H}}, \qquad \cos\theta = \frac{\text{A}}{\text{H}}, \qquad \tan\theta = \frac{\text{O}}{\text{A}} \]

The mnemonic SOHCAHTOA packs all three in: Sine is Opposite over Hypotenuse, Cosine is Adjacent over Hypotenuse, Tangent is Opposite over Adjacent. To decide which ratio to use, look at which two sides the question involves and pick the ratio that contains both.

Set the calculator to degrees

Before typing anything, check the screen shows DEG (not RAD or GRAD). A wrong mode is the single most common reason a trig answer comes out strange, and it is invisible unless you look.

Finding a missing side

When you know an angle and one side and want another side, choose the ratio linking the two sides, substitute, then rearrange.

Worked example

A right-angled triangle has hypotenuse 10 cm and an angle of \(35^\circ\). Find the side opposite that angle, to 2 decimal places.

1
You have the hypotenuse and want the opposite, so use sine: \(\sin\theta = \dfrac{\text{O}}{\text{H}}\).
2
Substitute: \(\sin 35^\circ = \dfrac{\text{O}}{10}\).
3
Rearrange: \(\text{O} = 10 \times \sin 35^\circ = 10 \times 0.5736\ldots\).
Opposite \(\approx 5.74\) cm

Finding a missing angle

When you know two sides and want the angle, form the correct ratio, then undo it with the inverse function, written \(\sin^{-1}\), \(\cos^{-1}\) or \(\tan^{-1}\) (usually the shift or 2nd key on the calculator).

Worked example

A right-angled triangle has an opposite side of 7 cm and a hypotenuse of 10 cm for angle \(\theta\). Find \(\theta\) to 1 decimal place.

1
You have the opposite and hypotenuse, so use sine: \(\sin\theta = \dfrac{\text{O}}{\text{H}}\).
2
Substitute: \(\sin\theta = \dfrac{7}{10} = 0.7\).
3
Take the inverse: \(\theta = \sin^{-1}(0.7) = 44.42\ldots^\circ\).
\(\theta \approx 44.4^\circ\)

Where this is assessed

Choosing between Pythagoras and a trig ratio, then computing the answer, is Criterion A (Knowing and understanding). Setting out the rule, the substitution and a labelled sketch is Criterion C (Communicating). Applying it to ladders, ramps and heights is Criterion D (Applying mathematics in real-life contexts). If a missing angle comes out above \(90^\circ\), you have likely put a side ratio in upside down.

Check yourself

1. The two shorter sides are 9 cm and 12 cm. Find the hypotenuse. +

Use Pythagoras: \(c^2 = 9^2 + 12^2 = 81 + 144 = 225\). So \(c = \sqrt{225} = \mathbf{15}\) cm.

2. An angle is \(40^\circ\) and its adjacent side is 6 cm. Find the opposite side, to 2 decimal places. +

Opposite and adjacent means tangent: \(\tan 40^\circ = \dfrac{\text{O}}{6}\). Rearranging, \(\text{O} = 6 \times \tan 40^\circ = 6 \times 0.8391\ldots \approx \mathbf{5.03}\) cm.

3. The opposite side is 5 cm and the adjacent side is 12 cm. Find the angle, to 1 decimal place. +

Opposite and adjacent means tangent: \(\tan\theta = \dfrac{5}{12} = 0.4166\ldots\). So \(\theta = \tan^{-1}(0.4166\ldots) \approx \mathbf{22.6^\circ}\).

Part of the Standard Maths library. Spotted an error or want a topic added? That feedback makes the notes better.