4. Sine Rule

Question 1: [4 marks]

Consider the following triangle

$\angle BAC = 30 \degree$

$\angle ABC = 80 \degree$

Find the length of $x$, to $2$ decimal places.

 

Answer type: Simple text answer

ANSWER: 2.54 cm

WORKING:

$\dfrac{x}{\sin(30)} = \dfrac{5}{\sin(80)}$

$x = \dfrac{5\sin(30)}{\sin(80)} = 2.54$ cm ($2$ dp)

 

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Question 2: [4 marks]

Consider the following triangle.

$\angle ABC = 33.1 \degree$

Calculate the angle $BCA$, to $2$ decimal places.

 

Answer type: Simple text answer

ANSWER: 22.96 $\degree$

WORKING:

$\dfrac{\sin(BCA)}{5} = \dfrac{\sin(33.1)}{7}$

$\sin(BCA) = \dfrac{5\sin(33.1)}{7}$

$BCA = \sin^{-1} \bigg(\dfrac{5\sin(33.1)}{7} \bigg) = 22.96$ ($2$ dp)

 

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Question 3: [4 marks]

The diagram below shows a triangle.

$\angle BAC = 15 \degree$

$x$ is an obtuse angle.

Work out the size of angle $x$, to $3$ significant figures.

 

Answer type: Simple text answer

ANSWER: 154 $\degree$

WORKING:

$\dfrac{\sin(x)}{12} = \dfrac{\sin(15)}{7}$

$\sin(x) = \dfrac{12\sin(15)}{7} = 0.4437...$

$x = \sin^{-1}(0.4437...) = 26.3395...$

Since $x$ is obtuse,

$x = 180 - 26.3396... = 154 \degree$ ($3$ sf)

 

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Question 4:

The diagram below shows a triangle.

$MO = 12$ cm

$LM = 6.5$ cm

$\angle OLM = 52 \degree$

 

Question 4(a): [3 marks]

Find the angle $MOL$, to $1$ decimal place.

 

Answer type: Simple text answer

ANSWER: 25.3 $\degree$

WORKING:

$\dfrac{\sin(MOL)}{6.5} = \dfrac{\sin(52)}{12}$

$\sin(MOL) = \dfrac{6.5 \sin(52)}{12}$

$MOL = \sin^{-1} \bigg(\dfrac{6.5 \sin(52)}{12} \bigg) = 25.3 \degree$

 

 

Question 4(b): [4 marks]

Find the length $LO$ using the sine rule, to $1$ decimal place.

 

Answer type: Simple text answer

ANSWER: 14.9 cm

WORKING:

$\angle LMO = 102.7 \degree$

$\dfrac{LO}{\sin(102.7)} = \dfrac{12}{\sin(52)}$

$LO = \dfrac{12 \sin(102.7)}{\sin(52)} = 14.9$ cm

 

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Question 5:

The diagram below shows a triangle with angles $x$, $50 \degree$ and $2x-35$.

 

Question 5(a): [1 mark]

Work out the value of $x$.

 

Answer type: Simple text answer

ANSWER: x = 55

WORKING:

$x + (2x-35) + 50 = 180$

$3x+15 = 180$

$x = 55$

 

 

Question 5(b): [4 marks]

Calculate the length $BC$, to $2$ decimal places.

 

Answer type: Simple text answer

ANSWER: 14.15 cm

WORKING:

$\angle BAC = 2(55) - 35 = 75 \degree$

$\dfrac{\sin(55)}{12} = \dfrac{\sin(75)}{BC}$

$BC = \dfrac{12 \sin(75)}{\sin(55)} = 14.15$ ($2$ dp)

 

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Question 6:

In the diagram,

$AD = CD = 6$ cm

$\angle ACD = 40 \degree$

$\angle ABD = 45 \degree$

 

Question 6(a): [3 marks]

Calculate the length $AC$, to $2$ significant figures.

 

Answer type: Simple text answer

ANSWER: 9.2 cm

WORKING:

$\angle ADC = 100 \degree$, since the triangle $ADC$ is isosceles.

$\dfrac{AC}{\sin(100)} = \dfrac{6}{\sin(40)}$

$AC = \dfrac{6 \sin(100)}{\sin(40)} = 9.2$ m ($2$ sf)

 

 

Question 6(b): [4 marks]

Calculate the length $BD$, to $2$ decimal places.

 

Answer type: Simple text answer

ANSWER: 6.95 cm

WORKING:

$\angle ADB = 180 - 100 = 80 \degree$

$\angle BAD = 180 - 80 - 45 = 55 \degree$

$\dfrac{BD}{\sin(55)} = \dfrac{6}{\sin(45)}$

$BD = \dfrac{6 \sin(55)}{\sin(45)} = 6.95$ m ($2$ dp)