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GCSE Edexcel Maths › Law of Sine and Cosine (GCSE Maths)

Law of Sine and Cosine (GCSE Maths)

Skill Check

Q1
Question 1

Work out the length of AC.

Give your answer to 1 decimal place.

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SOLUTION

Step 1  ·  Calculate the missing angle $B$

First, calculate the missing angle $B$ because side $AC$ is opposite angle $B$.

$$\angle B = 180^\circ - (55^\circ + 20^\circ) = 105^\circ$$

Step 2  ·  Use the sine rule to find side $AC$

Next, use the sine rule to find side $AC$.

$$\frac{AC}{\sin(105^\circ)} = \frac{12}{\sin(55^\circ)}$$
$$AC = \frac{12 \times \sin(105^\circ)}{\sin(55^\circ)}$$
$$AC \approx 14.15\dots\text{ cm}$$

Step 3  ·  Round the answer

Rounding to 3 significant figures gives:

Final Answer

$14.2\text{ cm}$

Q2
Question 2

Work out the size of angle x.

Give your answer to 3 significant figures.

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SOLUTION

Step 1  ·  Set up the sine rule equation

Using the sine rule $\left(\dfrac{\sin A}{a} = \dfrac{\sin B}{b}\right)$, substitute the given values to find angle $x$:

$$\frac{\sin(x)}{10} = \frac{\sin(60^\circ)}{15}$$

Step 2  ·  Rearrange to solve for $\sin(x)$

Multiply both sides by $10$ to isolate $\sin(x)$:

$$\sin(x) = \frac{10 \times \sin(60^\circ)}{15}$$
$$\sin(x) \approx 0.5773...$$

Step 3  ·  Calculate the inverse sine

Take the inverse sine ($\arcsin$ or $\sin^{-1}$) to find the angle $x$:

$$x = \arcsin(0.5773...)$$
$$x = 35.3^\circ$$

Final Answer

$x = 35.3^\circ$

Q3
Question 3

Work out the value of x.

Give your answer to 3 significant figures.

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SOLUTION

Step 1  ·  Set up the sine rule

Using the sine rule, we equate the ratio of the sine of an angle to its opposite side:

$$\frac{\sin(x)}{5.4} = \frac{\sin(95^\circ)}{6.7}$$

Step 2  ·  Rearrange to isolate $\sin(x)$

Multiply both sides by $5.4$ to get $\sin(x)$ on its own:

$$\sin(x) = \frac{5.4 \times \sin(95^\circ)}{6.7}$$
$$\sin(x) = 0.8029\dots$$

Step 3  ·  Calculate angle $x$

Use the inverse sine function ($\arcsin$ or $\sin^{-1}$) to find the angle:

$$x = \arcsin(0.8029\dots)$$

Final Answer

$x = 53.4^\circ$

Q4
Question 4

Work out the length of BC.

Give your answer to 3 significant figures.

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SOLUTION

Step 1  ·  Apply the sine rule

Use the sine rule to set up an equation for side $BC$:

$$\frac{BC}{\sin(42^\circ)} = \frac{5}{\sin(53^\circ)}$$
$$BC = \frac{5 \times \sin(42^\circ)}{\sin(53^\circ)}$$

Step 2  ·  Calculate the length

$$BC = 4.189\dots$$

Final Answer

$BC = 4.19\text{ m}$

Q5
Question 5

Work out the value of x.

Give your answer to 1 decimal place.

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SOLUTION

Step 1  ·  Set up the sine rule

Use the sine rule to find the missing side $x$:

$$\frac{x}{\sin(38^\circ)} = \frac{13}{\sin(100^\circ)}$$

Step 2  ·  Rearrange and calculate

$$x = \frac{13 \times \sin(38^\circ)}{\sin(100^\circ)}$$
$$x = 8.127\dots$$

Step 3  ·  Round the answer

Rounding to 1 decimal place gives:

Final Answer

$8.1\text{ cm}$

Problem Solving

Q6
Question 6

The area of the triangle is 100 m².

Calculate the perimeter of triangle ABC.

Give your answer to 3 significant figures.

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SOLUTION

Step 1  ·  Find angle $A$ using the area formula

Use the area formula $\text{Area} = \frac{1}{2}bc\sin(A)$ to find the angle between sides $19.7\text{ m}$ and $15.4\text{ m}$:

$$100 = \frac{1}{2} \times 19.7 \times 15.4 \times \sin(A)$$
$$100 = 151.69 \times \sin(A)$$
$$\sin(A) = \frac{100}{151.69} \approx 0.65924$$
$$A = \sin^{-1}(0.65924) \approx 41.23^\circ$$

Step 2  ·  Find the third side using the cosine rule

Use the cosine rule to find the length of the third side, $BC$:

$$BC^2 = 19.7^2 + 15.4^2 - (2 \times 19.7 \times 15.4 \times \cos(41.23^\circ))$$
$$BC^2 = 388.09 + 237.16 - (606.76 \times 0.7521)$$
$$BC^2 = 625.25 - 456.33 = 168.92$$
$$BC = \sqrt{168.92} \approx 13.00\text{ m}$$

Step 3  ·  Calculate the total perimeter

Add the lengths of all three sides together:

$$\text{Perimeter} = 19.7 + 15.4 + 13.00 = 48.1\text{ m}$$

Final Answer

$48.1\text{ m}$

Q7
Question 7

Work out the value of x.

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SOLUTION

Step 1  ·  Apply the cosine rule

Use the cosine rule with the sides $(x + 2)$, $(2x - 3)$, and $\sqrt{73}$, and the included angle $60^\circ$:

$$(\sqrt{73})^2 = (x + 2)^2 + (2x - 3)^2 - 2(x + 2)(2x - 3)\cos(60^\circ)$$

Step 2  ·  Expand and simplify

Substitute $\cos(60^\circ) = 0.5$:

$$73 = (x^2 + 4x + 4) + (4x^2 - 12x + 9) - (2x^2 + x - 6)$$

Combine like terms:

$$73 = 3x^2 - 7x + 19$$

Step 3  ·  Form a quadratic equation

Rearrange into standard quadratic form ($ax^2 + bx + c = 0$):

$$3x^2 - 7x - 54 = 0$$

Step 4  ·  Solve using the quadratic formula

Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$$x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(3)(-54)}}{2(3)}$$
$$x = \frac{7 \pm \sqrt{49 + 648}}{6} = \frac{7 \pm \sqrt{697}}{6}$$

Step 5  ·  Evaluate the valid value for $x$

Since lengths must be positive ($x + 2 > 0$):

$$x = \frac{7 + 26.40}{6} \approx 5.57$$

Final Answer

$x = 5.6$

Q8
Question 8
[1 marks]

Work out the value of x.

Give your answer to 1 decimal place.

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SOLUTION

Step 1  ·  Find the shared middle side $h$

Use the cosine rule on the left triangle:

$$h^2 = 10^2 + 13^2 - (2 \times 10 \times 13 \times \cos(40^\circ))$$
$$h^2 = 100 + 169 - (260 \times 0.7660)$$
$$h^2 = 269 - 199.17 = 69.83$$
$$h = \sqrt{69.83} \approx 8.356\text{ cm}$$

Step 2  ·  Find angle $x$ using trigonometry

Use trigonometry on the right-angled triangle to find angle $x$:

$$\cos(x^\circ) = \frac{5}{8.356} \approx 0.5983$$
$$x = \cos^{-1}(0.5983) \approx 53.3^\circ$$

Final Answer

$53.3^\circ$

Q9
Question 9

Work out the size of angle BAC.

Give your answer to 3 significant figures.

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SOLUTION

Step 1  ·  Apply the cosine rule

Use the cosine rule to set up the equation for angle $BAC$:

$$\cos(\angle BAC) = \frac{5.2^2 + 7.3^2 - 6.9^2}{2 \times 5.2 \times 7.3}$$

Step 2  ·  Simplify the expression

$$\cos(\angle BAC) = \frac{27.04 + 53.29 - 47.61}{75.92}$$
$$\cos(\angle BAC) = \frac{32.72}{75.92} \approx 0.43098$$

Step 3  ·  Calculate the inverse cosine

$$\angle BAC = \cos^{-1}(0.43098)$$

Final Answer

$64.5^\circ$

Q10
Question 10

Work out the value of x.

Give your answer to 1 decimal place.

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SOLUTION

Step 1  ·  Apply the cosine rule

Use the cosine rule to find the angle $x$:

$$\cos(x^\circ) = \frac{21^2 + 13^2 - 24^2}{2 \times 21 \times 13}$$

Step 2  ·  Simplify the expression

$$\cos(x^\circ) = \frac{441 + 169 - 576}{546}$$
$$\cos(x^\circ) = \frac{34}{546} \approx 0.06227$$

Step 3  ·  Calculate the inverse cosine

$$x = \cos^{-1}(0.06227)$$

Final Answer

$x = 86.4^\circ$

Exam-Style Questions

Q11
Question 11

Calculate the size of angle ABD.

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SOLUTION

Step 1  ·  Calculate shared side $BD$ using the sine rule

Apply the sine rule to $\triangle BCD$:

$$\frac{BD}{\sin(50^\circ)} = \frac{8.4}{\sin(68^\circ)}$$
$$BD = \frac{8.4 \times \sin(50^\circ)}{\sin(68^\circ)}$$
$$BD \approx 6.940 \text{ cm}$$

Step 2  ·  Find angle $ABD$ using the cosine rule

Now that all three sides of $\triangle ABD$ are known, apply the cosine rule to find $\angle ABD$:

$$\cos(\angle ABD) = \frac{AB^2 + BD^2 - AD^2}{2(AB)(BD)}$$
$$\cos(\angle ABD) = \frac{6.5^2 + 6.940^2 - 5^2}{2 \times 6.5 \times 6.940}$$
$$\cos(\angle ABD) = \frac{42.25 + 48.1636 - 25}{90.22}$$
$$\cos(\angle ABD) = \frac{65.4136}{90.22} \approx 0.7250$$

Step 3  ·  Calculate the final angle

Take the inverse cosine to find the angle:

$$\angle ABD = \cos^{-1}(0.7250)$$
$$\angle ABD \approx 43.5^\circ$$

Final Answer

$43.5^\circ$

Q12
Question 12

Two small boats are 24 m apart.

The angle of elevation of the boats to the top of a lighthouse are 20° and 34° respectively.

Calculate the height of the lighthouse.

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SOLUTION

Step 1  ·  Define the variables

Let the horizontal distance from the closer boat to the base of the lighthouse be $x$ and the height of the lighthouse be $h$.

Step 2  ·  Set up an equation for the closer boat

Using the closer boat, which forms a right-angled triangle with the lighthouse, set up an equation using the tangent ratio:

$$\tan(34^\circ) = \frac{h}{x} \implies x = \frac{h}{\tan(34^\circ)}$$

Step 3  ·  Set up an equation for the further boat

Using the further boat, which forms a larger right-angled triangle, set up a second equation:

$$\tan(20^\circ) = \frac{h}{24 + x} \implies 24 + x = \frac{h}{\tan(20^\circ)}$$

Step 4  ·  Substitute to eliminate $x$

Substitute the expression for $x$ from step 2 into the equation from step 3:

$$24 + \frac{h}{\tan(34^\circ)} = \frac{h}{\tan(20^\circ)}$$

Step 5  ·  Rearrange and solve for $h$

Rearrange the equation to factor out and solve for $h$:

$$24 = \frac{h}{\tan(20^\circ)} - \frac{h}{\tan(34^\circ)}$$
$$24 = h \left( \frac{1}{\tan(20^\circ)} - \frac{1}{\tan(34^\circ)} \right)$$
$$h = \frac{24}{\frac{1}{\tan(20^\circ)} - \frac{1}{\tan(34^\circ)}}$$

Final Answer

$18.97\text{ m}$

Q13
Question 13

In a quadrilateral ABCD, AD = 7 cm, AB = 8 cm and CD = 14 cm.

Angle BAD = 150° and Angle ADC = 70°.

Calculate the length BC.

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SOLUTION

Step 1  ·  Calculate the length of diagonal $BD$

By drawing a diagonal line from $B$ to $D$, we form $\triangle ABD$. Using the cosine rule to calculate the length of the diagonal $BD$:

$$BD^2 = 8^2 + 7^2 - 2(8)(7)\cos(150^\circ)$$
$$BD^2 = 64 + 49 - 112(-0.8660) \approx 209.99$$
$$BD = \sqrt{209.99} \approx 14.49\text{ cm}$$

Step 2  ·  Find angle $ADB$ using the sine rule

Use the sine rule in $\triangle ABD$ to find $\angle ADB$:

$$\frac{\sin(\angle ADB)}{8} = \frac{\sin(150^\circ)}{14.49}$$
$$\sin(\angle ADB) = \frac{8 \times 0.5}{14.49} \approx 0.2760$$
$$\angle ADB = \sin^{-1}(0.2760) \approx 16.02^\circ$$

Step 3  ·  Calculate angle $BDC$

Find $\angle BDC$ by subtracting $\angle ADB$ from the total $\angle ADC$:

$$\angle BDC = 70^\circ - 16.02^\circ = 53.98^\circ$$

Step 4  ·  Calculate the length of $BC$

Use the cosine rule in $\triangle BCD$ to calculate the length of $BC$:

$$BC^2 = 14.49^2 + 14^2 - 2(14.49)(14)\cos(53.98^\circ)$$
$$BC^2 \approx 209.96 + 196 - 405.72(0.5880) \approx 167.40$$
$$BC = \sqrt{167.40} \approx 12.94\text{ cm}$$

Final Answer

$12.94\text{ cm}$

Q14
Question 14

A boat, located at position X is running out of fuel.

There are two ports located at Y and Z.

The boat must refuel as soon as possible.

How much closer is the boat to the port at Y than the port at Z?

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SOLUTION

Step 1  ·  Find the missing angle at $X$

Calculate the missing angle at $X$ using the sum of angles in a triangle:

$$\angle X = 180^\circ - (60^\circ + 45^\circ) = 75^\circ$$

Step 2  ·  Calculate the distance to port $Y$

Use the sine rule to calculate the distance from the boat to port $Y$ (side $XY$):

$$\frac{XY}{\sin(45^\circ)} = \frac{28}{\sin(75^\circ)}$$
$$XY = \frac{28 \times \sin(45^\circ)}{\sin(75^\circ)}$$
$$XY \approx 20.50\text{ miles}$$

Step 3  ·  Calculate the distance to port $Z$

Use the sine rule to calculate the distance from the boat to port $Z$ (side $XZ$):

$$\frac{XZ}{\sin(60^\circ)} = \frac{28}{\sin(75^\circ)}$$
$$XZ = \frac{28 \times \sin(60^\circ)}{\sin(75^\circ)}$$
$$XZ \approx 25.10\text{ miles}$$

Step 4  ·  Find the difference in distance

Calculate the difference between the two distances to find how much closer the boat is to port $Y$:

$$\text{Difference} = 25.10 - 20.50$$
$$\text{Difference} \approx 4.60\text{ miles}$$

Final Answer

$4.60\text{ miles}$

Q15
Question 15

Shown below is triangle ABC.

Side AB is 40 cm, side BC is 22 cm, and angle C is 138°.

Calculate the length of AC.

Give your answer to 2 decimal places.

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SOLUTION

Step 1  ·  Find angle $A$ using the sine rule

$$\frac{\sin A}{22} = \frac{\sin 138^\circ}{40}$$
$$\sin A = \frac{22 \times \sin 138^\circ}{40} \approx 0.3680$$
$$A = \sin^{-1}(0.3680) \approx 21.595^\circ$$

Step 2  ·  Calculate angle $B$

Angles in a triangle add up to $180^\circ$:

$$B = 180^\circ - (138^\circ + 21.595^\circ)$$
$$B = 20.405^\circ$$

Step 3  ·  Calculate the length of side $AC$

Using the sine rule again:

$$\frac{AC}{\sin 20.405^\circ} = \frac{40}{\sin 138^\circ}$$
$$AC = \frac{40 \times \sin 20.405^\circ}{\sin 138^\circ}$$
$$AC \approx 20.84\text{ cm}$$

Final Answer

$20.84\text{ cm}$

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