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Trigonometry - Right-angled Trigonometry (SOHCAHTOA)

Grade 9IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Labeling a Right-Angled Triangle: Before applying SOHCAHTOA, identify the sides relative to the given angle θ\theta. The Hypotenuse (HH) is opposite the right angle, the Opposite (OO) is across from θ\theta, and the Adjacent (AA) is next to θ\theta.

A right-angled triangle with sides labeled Hypotenuse, Opposite, and Adjacent relative to angle theta.
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SOH: sin⁡(θ)=OppositeHypotenuse\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}}. Use this ratio when you are given or need to find the opposite side and the hypotenuse.

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CAH: cos⁡(θ)=AdjacentHypotenuse\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}}. Use this ratio when you are given or need to find the adjacent side and the hypotenuse.

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TOA: tan⁡(θ)=OppositeAdjacent\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}. Use this ratio when dealing with the two legs of the triangle (opposite and adjacent) without involving the hypotenuse.

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Inverse Trigonometry: When finding an unknown angle, use the inverse functions sin⁡−1\sin^{-1}, cos⁡−1\cos^{-1}, or tan⁡−1\tan^{-1} on your calculator.

Instructional text showing inverse tangent to find an angle.

📐Formulae

sin⁡(θ)=OppositeHypotenuse\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}}

cos⁡(θ)=AdjacentHypotenuse\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}}

tan⁡(θ)=OppositeAdjacent\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}

θ=sin⁡−1(OppHyp)\theta = \sin^{-1}\left(\frac{\text{Opp}}{\text{Hyp}}\right)

θ=cos⁡−1(AdjHyp)\theta = \cos^{-1}\left(\frac{\text{Adj}}{\text{Hyp}}\right)

θ=tan⁡−1(OppAdj)\theta = \tan^{-1}\left(\frac{\text{Opp}}{\text{Adj}}\right)

💡Examples

Problem 1:

In a right-angled triangle, the angle θ\theta is 32∘32^\circ and the hypotenuse is 1515 cm. Find the length of the opposite side xx.

Solution:

x=15×sin⁡(32∘)≈7.95x = 15 \times \sin(32^\circ) \approx 7.95 cm

Explanation:

Identify the knowns: Hypotenuse = 15, Angle = 32°. We need the Opposite side. SOH tells us to use Sine. sin⁡(32∘)=x15\sin(32^\circ) = \frac{x}{15}. Multiplying both sides by 15 gives x=15sin⁡(32∘)x = 15 \sin(32^\circ).

Problem 2:

A ladder 55 m long leans against a vertical wall. The base of the ladder is 33 m away from the wall. Calculate the angle the ladder makes with the ground.

Solution:

θ=cos⁡−1(35)=53.1∘\theta = \cos^{-1}(\frac{3}{5}) = 53.1^\circ

Explanation:

The ladder represents the Hypotenuse (5m) and the distance from the wall is the Adjacent side (3m). CAH tells us to use Cosine: cos⁡(θ)=35\cos(\theta) = \frac{3}{5}. To find the angle, use the inverse cosine function.

Problem 3:

Find the length of the adjacent side if the opposite side is 88 cm and the angle is 40∘40^\circ.

Solution:

Adjacent=8tan⁡(40∘)≈9.53\text{Adjacent} = \frac{8}{\tan(40^\circ)} \approx 9.53 cm

Explanation:

We have the opposite side and need the adjacent side. TOA tells us to use Tangent. tan⁡(40∘)=8adj\tan(40^\circ) = \frac{8}{\text{adj}}. Rearranging for 'adj' gives adj=8tan⁡(40∘)\text{adj} = \frac{8}{\tan(40^\circ)}.

Problem 4:

A ramp is 1212 m long and makes an angle of 15∘15^\circ with the horizontal ground. Calculate the vertical height hh of the ramp.

Diagram of a ramp forming a right-angled triangle with height h, length 12m, and angle 15 degrees.

Solution:

sin⁡(15∘)=h12\sin(15^\circ) = \frac{h}{12} h=12×sin⁡(15∘)h = 12 \times \sin(15^\circ) h=12×0.2588...h = 12 \times 0.2588... h≈3.11 m (2 d.p.)h \approx 3.11 \text{ m (2 d.p.)}

Explanation:

We are given the hypotenuse (1212 m) and the angle (15∘15^\circ). We need to find the opposite side (hh). Therefore, we use the Sine ratio (SOH).

Problem 5:

A flagpole casts a shadow 77 m long on the ground. If the flagpole is 1010 m tall, calculate the angle of elevation xx of the sun.

Diagram showing a vertical pole of 10m, a horizontal shadow of 7m, and an angle of elevation x.

Solution:

tan⁡(x)=107\tan(x) = \frac{10}{7} x=tan⁡−1(107)x = \tan^{-1}\left(\frac{10}{7}\right) x=tan⁡−1(1.428...)x = \tan^{-1}(1.428...) x≈55.0∘ (1 d.p.)x \approx 55.0^\circ \text{ (1 d.p.)}

Explanation:

We are given the opposite side (height of pole = 1010 m) and the adjacent side (shadow length = 77 m). We need to find the angle xx, so we use the inverse Tangent ratio (TOA).