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Geometry and Measurement - Trigonometric Ratios in Right-Angled Triangles

Grade 8IB

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

🔑Concepts

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In a right-angled triangle, the three sides are named relative to a specific acute angle θ\theta: the Hypotenuse (longest side across from the 90∘90^{\circ} angle), the Opposite side (across from θ\theta), and the Adjacent side (next to θ\theta).

A right-angled triangle showing the relative positions of the hypotenuse, opposite, and adjacent sides to angle theta.
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The mnemonic SOH CAH TOA is used to remember the three primary ratios: Sine (Opposite/Hypotenuse), Cosine (Adjacent/Hypotenuse), and Tangent (Opposite/Adjacent).

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When two sides are known, the inverse trigonometric functions sin⁡−1\sin^{-1}, cos⁡−1\cos^{-1}, and tan⁡−1\tan^{-1} are used to calculate the measure of the unknown angle.

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Trigonometric ratios are constant for a given angle regardless of the triangle's size, reflecting the property of similarity in geometry.

📐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}}

a2+b2=c2 (Pythagorean Theorem)a^2 + b^2 = c^2 \text{ (Pythagorean Theorem)}

💡Examples

Problem 1:

In a right-angled triangle, the hypotenuse is 15 cm15 \text{ cm} and one of the acute angles is 30∘30^{\circ}. Calculate the length of the side opposite to the 30∘30^{\circ} angle.

Solution:

sin⁡30∘=x15\sin 30^{\circ} = \frac{x}{15} 0.5=x150.5 = \frac{x}{15} x=15×0.5x = 15 \times 0.5 x=7.5 cmx = 7.5 \text{ cm}

Explanation:

Since we are given the hypotenuse and the angle, and we need to find the opposite side, we use the Sine ratio (SOHSOH). Multiply the hypotenuse by sin⁡30∘\sin 30^{\circ} to find the length.

Problem 2:

A right-angled triangle has an adjacent side of 8 cm8 \text{ cm} and an opposite side of 6 cm6 \text{ cm} relative to an angle θ\theta. Find the value of θ\theta.

Solution:

tan⁡θ=68\tan \theta = \frac{6}{8} tan⁡θ=0.75\tan \theta = 0.75 θ=tan⁡−1(0.75)\theta = \tan^{-1}(0.75) θ≈36.87∘\theta \approx 36.87^{\circ}

Explanation:

We are given the opposite and adjacent sides, so we use the Tangent ratio (TOATOA). To find the angle, we apply the inverse tangent function tan⁡−1\tan^{-1} to the ratio of the sides.

Problem 3:

Find the length of the hypotenuse hh if the adjacent side is 10 cm10 \text{ cm} and the angle θ\theta is 60∘60^{\circ}.

Solution:

cos⁡60∘=10h\cos 60^{\circ} = \frac{10}{h} 0.5=10h0.5 = \frac{10}{h} h=100.5h = \frac{10}{0.5} h=20 cmh = 20 \text{ cm}

Explanation:

We use the Cosine ratio (CAHCAH) because we are dealing with the adjacent side and the hypotenuse. Rearranging the formula h=Adjacentcos⁡θh = \frac{\text{Adjacent}}{\cos \theta} gives the result.

Problem 4:

A ladder leans against a vertical wall. The foot of the ladder is 5 m5 \text{ m} away from the wall, and the ladder makes an angle of 65∘65^{\circ} with the ground. How high up the wall does the ladder reach?

Diagram showing a ladder forming a right-angled triangle with a wall and the ground.

Solution:

tan⁡65∘=OppositeAdjacent\tan 65^{\circ} = \frac{\text{Opposite}}{\text{Adjacent}} tan⁡65∘=h5\tan 65^{\circ} = \frac{h}{5} h=5×tan⁡65∘h = 5 \times \tan 65^{\circ} h≈5×2.1445h \approx 5 \times 2.1445 h≈10.72 mh \approx 10.72 \text{ m}

Explanation:

We use the tangent ratio because we are given the adjacent side (5 m5 \text{ m}) and need to find the opposite side (height hh). Substituting the values and solving for hh gives the height reached by the ladder.

Problem 5:

A slide is 12 m12 \text{ m} long. If the top of the slide is 7 m7 \text{ m} above the ground, what angle α\alpha does the slide make with the ground?

Diagram of a slide forming the hypotenuse of a right-angled triangle with height 7m and length 12m.

Solution:

sin⁡α=OppositeHypotenuse\sin \alpha = \frac{\text{Opposite}}{\text{Hypotenuse}} sin⁡α=712\sin \alpha = \frac{7}{12} α=sin⁡−1(712)\alpha = \sin^{-1}\left(\frac{7}{12}\right) α≈sin⁡−1(0.5833)\alpha \approx \sin^{-1}(0.5833) α≈35.69∘\alpha \approx 35.69^{\circ}

Explanation:

To find the angle, we identify that the vertical height is the 'Opposite' side and the length of the slide is the 'Hypotenuse'. Using the sine ratio and the inverse sine function, we calculate the angle.