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

Grade 10IB

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

🔑Concepts

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Labelling a Right-Angled Triangle: Every right-angled triangle consists of a 90∘90^{\circ} angle, usually denoted by a small square in the corner. The side opposite this right angle is the 'Hypotenuse', which is always the longest side. When focusing on a specific reference angle θ\theta, the side across from it is the 'Opposite' side, and the side next to it (forming the angle with the hypotenuse) is the 'Adjacent' side.

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The SOH CAH TOA Mnemonic: This is a memory aid used to define the three primary trigonometric ratios. 'SOH' stands for sin⁡(θ)=OppositeHypotenuse\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}}, 'CAH' stands for cos⁡(θ)=AdjacentHypotenuse\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}}, and 'TOA' stands for tan⁡(θ)=OppositeAdjacent\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}. Visually, these ratios represent the constant relationship between the side lengths for any given angle θ\theta.

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Finding Unknown Side Lengths: To find a missing side, identify the known angle and the side you have, then choose the ratio that includes the side you want to find. If the unknown is in the numerator, multiply the denominator by the trigonometric value (e.g., x=H×sin⁡(θ)x = H \times \sin(\theta)). If the unknown is in the denominator, swap it with the trigonometric function (e.g., H=Osin⁡(θ)H = \frac{O}{\sin(\theta)}).

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Finding Unknown Angles using Inverse Ratios: When two sides of a right-angled triangle are known, the reference angle θ\theta can be found using the inverse trigonometric functions: sin⁡−1\sin^{-1}, cos⁡−1\cos^{-1}, and tan⁡−1\tan^{-1}. For example, if you know the Opposite and Adjacent sides, you calculate θ=tan⁡−1(OppositeAdjacent)\theta = \tan^{-1}(\frac{\text{Opposite}}{\text{Adjacent}}). This operation 'undoes' the trig function to isolate the angle.

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Angle of Elevation and Depression: These are angles measured from a horizontal line of sight. The Angle of Elevation is the angle looking 'upward' from the horizontal to an object. The Angle of Depression is the angle looking 'downward' from the horizontal to an object. Visually, if you draw a horizontal line from the observer's eye, the angle of depression to an object below is equal to the angle of elevation from that object back to the observer, due to alternate interior angles.

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The Pythagorean Theorem: Before or after using trigonometry, you may need to find a third side using a2+b2=c2a^2 + b^2 = c^2, where cc is the hypotenuse. This relationship is specific to right-angled triangles and works independently of the internal angles (other than the 90∘90^{\circ} angle).

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Calculator Settings: It is crucial for IB students to ensure their calculator is set to 'Degree' mode (DEGDEG) rather than 'Radian' mode (RADRAD) when solving standard geometric problems, as most Grade 10 problems provide angles in degrees.

📐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)

a2+b2=c2a^2 + b^2 = c^2

💡Examples

Problem 1:

A 6-meter ladder leans against a vertical wall. The ladder makes an angle of 70∘70^{\circ} with the ground. How far up the wall does the ladder reach?

Solution:

  1. Identify the given values: Hypotenuse (ladder length) = 6 m6\text{ m}, Angle θ=70∘\theta = 70^{\circ}. We need to find the Opposite side (height of the wall, hh).
  2. Choose the ratio: Since we have Hypotenuse and want Opposite, use SINE.
  3. Set up the equation: sin⁡(70∘)=h6\sin(70^{\circ}) = \frac{h}{6}
  4. Rearrange to solve for hh: h=6×sin⁡(70∘)h = 6 \times \sin(70^{\circ})
  5. Calculate: h≈6×0.9397=5.64 mh \approx 6 \times 0.9397 = 5.64\text{ m} (to 3 significant figures).

Explanation:

This problem uses the SOH ratio. Because the unknown height is the 'Opposite' side relative to the ground angle, we multiply the hypotenuse by the sine of the angle.

Problem 2:

A rectangular field is 15 m15\text{ m} long and 10 m10\text{ m} wide. Find the angle that the diagonal makes with the longer side.

Solution:

  1. Identify the triangle: The diagonal splits the rectangle into two right-angled triangles. For the angle θ\theta between the diagonal and the 15 m15\text{ m} side, the Adjacent side is 15 m15\text{ m} and the Opposite side is 10 m10\text{ m}.
  2. Choose the ratio: Since we have Opposite and Adjacent, use TANGENT.
  3. Set up the equation: tan⁡(θ)=1015\tan(\theta) = \frac{10}{15}
  4. Use the inverse function: θ=tan⁡−1(1015)\theta = \tan^{-1}\left(\frac{10}{15}\right)
  5. Calculate: θ=tan⁡−1(0.6667)≈33.7∘\theta = \tan^{-1}(0.6667) \approx 33.7^{\circ} (to 3 significant figures).

Explanation:

To find an unknown angle when two sides are known, we identify the relevant trig ratio (TOA in this case) and use the inverse trigonometric function on the calculator.