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)
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)
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)
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
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)
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)