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Geometry and Trigonometry - Right-angled triangle trigonometry (SOH CAH TOA)

Grade 9IB

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

🔑Concepts

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The naming of sides in a right-angled triangle depends on the position of the reference angle θ\theta. The Hypotenuse is always opposite the right angle. The Opposite side is across from θ\theta, and the Adjacent side is next to θ\theta.

A right-angled triangle with sides labeled Hypotenuse, Opposite, and Adjacent relative to angle theta.
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The mnemonic SOH CAH TOA helps remember the primary trigonometric ratios: sin⁡(θ)=OH\sin(\theta) = \frac{O}{H}, cos⁡(θ)=AH\cos(\theta) = \frac{A}{H}, and tan⁡(θ)=OA\tan(\theta) = \frac{O}{A}.

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To find an unknown side length, identify the two known values (one angle and one side) and use the ratio that connects them to the unknown side.

Diagram showing a triangle with hypotenuse 10 and angle 30 degrees to find side x.
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To find an unknown angle, use the inverse trigonometric functions sin⁡−1\sin^{-1}, cos⁡−1\cos^{-1}, or tan⁡−1\tan^{-1} when two side lengths are known.

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The Angle of Elevation is the angle measured upwards from the horizontal line of sight to an object.

Diagram of angle of elevation relative to a horizontal line.

📐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)

θ=cos⁡−1(AdjacentHypotenuse)\theta = \cos^{-1}\left(\frac{\text{Adjacent}}{\text{Hypotenuse}}\right)

θ=tan⁡−1(OppositeAdjacent)\theta = \tan^{-1}\left(\frac{\text{Opposite}}{\text{Adjacent}}\right)

a2+b2=c2a^2 + b^2 = c^2 (Pythagorean Theorem)

💡Examples

Problem 1:

In a right-angled triangle, the hypotenuse is 1212 cm long and one of the acute angles is 35∘35^\circ. Find the length of the side opposite to the 35∘35^\circ angle, correct to 2 decimal places.

Solution:

  1. Identify the given information: Hypotenuse =12= 12, θ=35∘\theta = 35^\circ, and we need to find the Opposite side (xx).
  2. Choose the correct ratio: Since we have the Hypotenuse and want the Opposite, we use SOH: sin⁡(θ)=OppositeHypotenuse\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}}.
  3. Set up the equation: sin⁡(35∘)=x12\sin(35^\circ) = \frac{x}{12}.
  4. Rearrange to solve for xx: x=12×sin⁡(35∘)x = 12 \times \sin(35^\circ).
  5. Calculate: x≈12×0.5736≈6.88x \approx 12 \times 0.5736 \approx 6.88 cm.

Explanation:

We use the Sine ratio because the problem involves the hypotenuse and the side opposite the given angle. Multiplying the hypotenuse by the sine of the angle isolates the unknown side length.

Problem 2:

A ladder is leaning against a wall. The foot of the ladder is 33 m away from the base of the wall, and the ladder reaches 77 m up the wall. Calculate the angle that the ladder makes with the ground.

Solution:

  1. Identify the given information: The distance from the wall is the Adjacent side (33 m) and the height up the wall is the Opposite side (77 m). We need to find the angle θ\theta.
  2. Choose the correct ratio: Since we have Opposite and Adjacent, we use TOA: tan⁡(θ)=OppositeAdjacent\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}.
  3. Set up the equation: tan⁡(θ)=73\tan(\theta) = \frac{7}{3}.
  4. Use the inverse tangent function to find θ\theta: θ=tan⁡−1(73)\theta = \tan^{-1}\left(\frac{7}{3}\right).
  5. Calculate: θ≈tan⁡−1(2.333)≈66.8∘\theta \approx \tan^{-1}(2.333) \approx 66.8^\circ.

Explanation:

Because we know the two legs of the triangle (opposite and adjacent) but not the hypotenuse, the tangent ratio is the most direct way to find the angle. We use the inverse tangent function to convert the ratio of the sides back into an angle measurement.

Problem 3:

A surveyor stands 5050 m from the base of a tower. The angle of elevation to the top of the tower is 28∘28^\circ. Calculate the height of the tower to the nearest meter.

Triangle representing a tower height calculation with distance 50m and angle 28 degrees.

Solution:

  1. Identify knowns: Adjacent =50= 50 m, θ=28∘\theta = 28^\circ.
  2. Identify unknown: Opposite (height hh).
  3. Use the tangent ratio: tan⁡(28∘)=h50\tan(28^\circ) = \frac{h}{50}
  4. Rearrange to solve for hh: h=50×tan⁡(28∘)h = 50 \times \tan(28^\circ)
  5. Calculate: h≈50×0.5317=26.585h \approx 50 \times 0.5317 = 26.585
  6. Round to the nearest meter: 2727 m.

Explanation:

We use the tangent ratio because we are relating the 'Opposite' side (height) and the 'Adjacent' side (distance from the base).

Problem 4:

A string of a kite is 8080 m long and makes an angle of 55∘55^\circ with the horizontal ground. Determine the vertical height of the kite above the ground, assuming the string is taut.

Kite diagram showing a string of 80m at a 55 degree angle to calculate height y.

Solution:

  1. Identify knowns: Hypotenuse =80= 80 m, θ=55∘\theta = 55^\circ.
  2. Identify unknown: Opposite (vertical height yy).
  3. Use the sine ratio: sin⁡(55∘)=y80\sin(55^\circ) = \frac{y}{80}
  4. Rearrange to solve for yy: y=80×sin⁡(55∘)y = 80 \times \sin(55^\circ)
  5. Calculate: y≈80×0.8192=65.536y \approx 80 \times 0.8192 = 65.536
  6. The height is 65.565.5 m (to 1 decimal place).

Explanation:

Since we know the length of the string (hypotenuse) and want to find the height (opposite), the sine ratio is the correct choice.