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GCSE Edexcel Maths › SOHCAHTOA (GCSE Maths)

SOHCAHTOA (GCSE Maths)

Skill Check

Q1
Question 1

Find the size of the missing angles in the triangles below.

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SOLUTION

Step 1  ·  Identify the given information

Relative to angle $x$, we have the following lengths:

  • The adjacent side is $20\text{ cm}$.
  • The hypotenuse is $25\text{ cm}$.

Step 2  ·  Select the trigonometric ratio

Use the cosine ratio, which relates the adjacent side to the hypotenuse:

$$\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}}$$

Step 3  ·  Set up and solve the equation

Substitute the known values and solve for $x$ using the inverse cosine function:

$$\cos(x) = \frac{20}{25}$$
$$x = \cos^{-1}\left(0.8\right) \approx 36.9^\circ$$

Final Answer

$x = 36.9^\circ$

Q2
Question 2

Find the size of the missing angles in the triangles below.

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SOLUTION

Step 1  ·  Identify the sides relative to angle $x$

  • The adjacent side to angle $x$ is $11\text{ cm}$.
  • The hypotenuse of the right-angled triangle is $15\text{ cm}$.

Step 2  ·  Set up the trigonometric equation

Use the cosine ratio, which is $\cos(\theta) = \dfrac{\text{Adjacent}}{\text{Hypotenuse}}$:

$$\cos(x) = \frac{11}{15}$$

Step 3  ·  Solve for $x$ using inverse cosine

$$x = \cos^{-1}\left(\frac{11}{15}\right) \approx 42.8^\circ$$

Final Answer

$x = 42.8^\circ$

Q3
Question 3

Find the lengths of the sides labelled x below.

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SOLUTION

Step 1  ·  Identify the given information

  • Relative to the $45^\circ$ angle, the adjacent side is $x$ and the hypotenuse is $1.3\text{ m}$.
  • Use the cosine trigonometric ratio, which is $\frac{\text{Adjacent}}{\text{Hypotenuse}}$.

Step 2  ·  Set up the equation and solve for $x$

$$\cos(45^\circ) = \frac{x}{1.3}$$
$$x = 1.3 \times \cos(45^\circ) = 0.919\text{ m}$$

Final Answer

$0.919\text{ m}$

Q4
Question 4

Find the lengths of the sides labelled x below.

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SOLUTION

Step 1  ·  Identify the given information

Relative to the $60^\circ$ angle, the opposite side is $x$ and the hypotenuse is $4\text{ km}$.

Step 2  ·  Set up the trigonometric equation

Use the sine ratio, which is opposite over hypotenuse:

$$\sin(60^\circ) = \frac{x}{4}$$

Step 3  ·  Solve for $x$

$$x = 4 \times \sin(60^\circ) = 3.46\text{ km}$$

Final Answer

$3.46\text{ km}$

Problem Solving

Q5
Question 5

Find the size of the missing angles in the triangles below.

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SOLUTION

Step 1  ·  Identify the given sides

Relative to angle $x$:

  • $\text{Opposite} = 7.4\text{ cm}$
  • $\text{Hypotenuse} = 9\text{ cm}$

Step 2  ·  Set up the trigonometric equation

Using the sine ratio $\left(\sin \theta = \dfrac{\text{Opposite}}{\text{Hypotenuse}}\right)$:

$$\sin(x) = \frac{7.4}{9}$$

Step 3  ·  Calculate angle $x$

Take the inverse sine $\left(\sin^{-1}\right)$ to find $x$:

$$x = \sin^{-1}\left(\frac{7.4}{9}\right)$$
$$x \approx 55.309\dots^\circ$$

Final Answer

$x = 55.3^\circ$

Q6
Question 6

Find the size of the missing angles in the triangles below.

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SOLUTION

Step 1  ·  Identify the given information

Relative to angle $x$, the opposite side is $2.5\text{ cm}$ and the adjacent side is $2.8\text{ cm}$.

Step 2  ·  Choose the trigonometric ratio

Use the tangent ratio, which is opposite over adjacent:

$$\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}$$

Step 3  ·  Calculate angle $x$

Set up the equation and solve using the inverse tangent function:

$$\tan(x) = \frac{2.5}{2.8}$$
$$x = \tan^{-1}\left(\frac{2.5}{2.8}\right) \approx 41.8^\circ$$

Final Answer

$41.8^\circ$

Q7
Question 7

Find the lengths of the sides labelled x below.

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SOLUTION

Step 1  ·  Identify the knowns and unknowns

  • Relative to the $48^\circ$ angle, the adjacent side is $28\text{ cm}$.
  • The hypotenuse is $x$.

Step 2  ·  Choose the correct trigonometric ratio

We need the ratio that connects the adjacent side and the hypotenuse, which is cosine:

$$\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}}$$

Step 3  ·  Set up the equation and solve for $x$

Substitute the known values into the ratio and rearrange to find $x$:

$$\cos(48^\circ) = \frac{28}{x}$$
$$x = \frac{28}{\cos(48^\circ)}$$

Final Answer

$x = 41.8\text{ cm}$

Exam-Style Questions

Q8
Question 8

Find the size of the missing angles in the triangles below.

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SOLUTION

Step 1  ·  Standardise the units

Convert $1.2\text{ m}$ to $120\text{ cm}$ so all measurements are in the same unit.

Step 2  ·  Identify the sides relative to angle $x$

  • The side opposite angle $x$ is $90\text{ cm}$.
  • The side adjacent to angle $x$ is $120\text{ cm}$.

Step 3  ·  Use the tangent ratio to find angle $x$

The tangent ratio is $\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}$. Setting up the equation gives:

$$\tan(x) = \frac{90}{120}$$
$$x = \tan^{-1}(0.75) \approx 36.9^\circ$$

Final Answer

$36.9^\circ$

Q9
Question 9

Find the lengths of the sides labelled x below.

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SOLUTION

Step 1  ·  Identify the known values and the correct trigonometric ratio

Relative to the $28^\circ$ angle:

  • The adjacent side is $17\text{ cm}$.
  • The opposite side is $x$.
  • This means we must use the tangent ratio $\left(\tan \theta = \dfrac{\text{Opposite}}{\text{Adjacent}}\right)$.

Step 2  ·  Set up the equation and solve for $x$

$$\tan(28^\circ) = \frac{x}{17}$$
$$x = 17 \times \tan(28^\circ)$$

Step 3  ·  Calculate the final value

$$x \approx 9.04\text{ cm}$$

Final Answer

$9.04\text{ cm}$

Q10
Question 10

Find the lengths of the sides labelled x below.

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SOLUTION

Step 1  ·  Identify the sides and choose a ratio

  • Relative to the $60^\circ$ angle, the opposite side is $x$.
  • The hypotenuse is $50\text{ cm}$.
  • This requires the sine trigonometric ratio, which is $\dfrac{\text{Opposite}}{\text{Hypotenuse}}$.

Step 2  ·  Set up the equation and solve for $x$

Substitute the known values into the sine ratio:

$$\sin(60^\circ) = \frac{x}{50}$$

Multiply both sides by 50 to isolate $x$:

$$x = 50 \times \sin(60^\circ) \approx 43.301\dots$$

Step 3  ·  State the final answer

Rounding to 1 decimal place gives:

Final Answer

$43.3\text{ cm}$

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