krit.club logo

Trigonometry - Sine, Cosine, and Tangent Ratios

Grade 10IGCSE

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

🔑Concepts

•

Labeling a Right-Angled Triangle: The Hypotenuse (longest side), Opposite (side across from the given angle), and Adjacent (side next to the given angle).

•

The SOH CAH TOA Mnemonic: A memory aid for the three primary ratios.

•

Finding Side Lengths: Using an angle and one known side to calculate an unknown side.

•

Finding Angles: Using the inverse trigonometric functions (arcsin, arccos, arctan) when two sides are known.

•

Angles of Elevation and Depression: Measured from the horizontal line of sight.

📐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(OppositeHypotenuse)\theta = \sin^{-1}\left(\frac{\text{Opposite}}{\text{Hypotenuse}}\right)

a2+b2=c2 (Pythagoras’ Theorem for related calculations)a^2 + b^2 = c^2 \text{ (Pythagoras' Theorem for related calculations)}

💡Examples

Problem 1:

In a right-angled triangle ABCABC, the angle BB is 90∘90^\circ, angle AA is 35∘35^\circ, and the hypotenuse ACAC is 1212 cm. Calculate the length of the side BCBC (Opposite to angle AA).

Solution:

BC=12×sin⁡(35∘)≈6.88BC = 12 \times \sin(35^\circ) \approx 6.88 cm

Explanation:

Identify that we have the hypotenuse and need the opposite side relative to the 35∘35^\circ angle. We use the Sine ratio: sin⁡(35∘)=BC12\sin(35^\circ) = \frac{BC}{12}. Rearranging gives BC=12sin⁡(35∘)BC = 12 \sin(35^\circ).

Problem 2:

A ladder 55 m long leans against a vertical wall. The foot of the ladder is 22 m away from the base of the wall. Find the angle the ladder makes with the ground.

Solution:

θ=cos⁡−1(25)≈66.4∘\theta = \cos^{-1}(\frac{2}{5}) \approx 66.4^\circ

Explanation:

The ladder is the hypotenuse (55 m) and the distance from the wall is the adjacent side (22 m). Since we have Adjacent and Hypotenuse, we use the Cosine ratio: cos⁡(θ)=25\cos(\theta) = \frac{2}{5}. Applying the inverse cosine gives the angle.

Problem 3:

Find the height of a flagpole if the angle of elevation from a point 1515 m away from its base to the top is 40∘40^\circ.

Solution:

Height=15×tan⁡(40∘)≈12.59\text{Height} = 15 \times \tan(40^\circ) \approx 12.59 m

Explanation:

The distance from the base is the adjacent side (1515 m) and the height is the opposite side. We use the Tangent ratio: tan⁡(40∘)=Height15\tan(40^\circ) = \frac{\text{Height}}{15}. Multiplying both sides by 1515 gives the height.