Question 1:
Choose the correct answer for each of the following calculations:
Question 1(a): [2 marks]
\cos(30 \degree) + \sin (60 \degree)
Answer type: Multiple choice type 1
A: \sqrt{3}
B: 1
C: 2
D: 2\sqrt{3}
ANSWER: A: \sqrt{3}
WORKING:
\dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{3}}{2} = \sqrt{3}
Question 1(b): [2 marks]
12 \cos (60 \degree) - 8 \sin(30 \degree)
Answer type: Multiple choice type 1
A: 2
B: 1
C: 2\sqrt{3}
D: \sqrt{3}
ANSWER: A: 2
WORKING:
12 \bigg(\dfrac{1}{2} \bigg) - 8 \bigg(\dfrac{1}{2} \bigg) = 6 - 4 = 2
Question 1(c): [2 marks]
\dfrac{\tan(45 \degree)}{\sin(30 \degree)} \times 10 \tan(60 \degree)
Answer type: Multiple choice type 1
A: 20 \sqrt{3}
B: 10 \sqrt{2}
C: 5\sqrt{3}
D: 2\sqrt{3}
ANSWER: A: 20 \sqrt{3}
WORKING:
\dfrac{1}{0.5} \times 10 \times \sqrt{3} = 20 \sqrt{3}
Question 1(d): [2 marks]
\tan(30 \degree) + \sin(60 \degree)
Answer type: Multiple choice type 1
A: \dfrac{5\sqrt{3}}{6}
B: \dfrac{1}{2} + \sqrt{3}
C: \dfrac{6\sqrt{3}}{5}
D: 2
ANSWER: A: \dfrac{5\sqrt{3}}{6}
WORKING:
\dfrac{1}{\sqrt{3}} + \dfrac{\sqrt{3}}{2} = \dfrac{2}{2\sqrt{3}} + \dfrac{3}{2\sqrt{3}} = \dfrac{5}{2\sqrt{3}} = \dfrac{5\sqrt{3}}{6}
Question 2:
Choose the correct answer for each of the following calculations:
Question 2(a): [2 marks]
\tan (30 \degree) + \sin(30 \degree)
Answer type: Multiple choice type 1
A: \dfrac{3+2\sqrt{3}}{6}
B: 2 - \sqrt{3}
C: 1
D: 2
ANSWER: A: \dfrac{3+2\sqrt{3}}{6}
WORKING:
\dfrac{1}{\sqrt{3}} + \dfrac{1}{2} = \dfrac{2}{2\sqrt{3}} + \dfrac{\sqrt{3}}{2\sqrt{3}} = \dfrac{2+\sqrt{3}}{2\sqrt{3}} = \dfrac{3+2\sqrt{3}}{6}
Question 2(b): [3 marks]
\dfrac{\tan (45 \degree) + \sin(30 \degree)}{\tan (60 \degree)} \times \cos (45 \degree)
Answer type: Multiple choice type 1
A: \dfrac{\sqrt{6}}{4}
B: \dfrac{\sqrt{3}+\sqrt{2}}{6}
C: \dfrac{\sqrt{6}}{2}
D: \dfrac{2\sqrt{6}+3\sqrt{2}}{12}
ANSWER: A: \dfrac{\sqrt{6}}{4}
WORKING:
\dfrac{\bigg(1+\dfrac{1}{2} \bigg)}{\sqrt{3}} \times \dfrac{\sqrt{2}}{2} = \dfrac{\sqrt{3}}{2} \times \dfrac{\sqrt{2}}{2} = \dfrac{\sqrt{6}}{4}
Question 3: [2 marks]
ABC is a right-angled triangle.
\angle BAC = 30 \degreeAB = 16 cm
Find the exact value of x, and choose the correct answer from below.
Answer type: Multiple choice type 1
A: 8 cm
B: 8\sqrt{3} cm
C: 16 \sqrt{3}
D: \dfrac{16 \sqrt{3}}{3} cm
ANSWER: A: 8 cm
WORKING:
x = 16 \sin(30 \degree) = 16 \times \dfrac{1}{2} = 8 cm
Question 4: [2 marks]
DEF is a right-angled triangle.
DE = EF
\angle EDF = 45 \degree
Find the exact value of x, and choose the correct answer from below.
Answer type: Multiple choice type 1
A: 5\sqrt{2} cm
B: 5\sqrt{3} cm
C: 5 cm
D: \dfrac{\sqrt{2}}{2} cm
ANSWER: A: 5\sqrt{2}
WORKING:
x = 10 \times \cos(45 \degree) = 10 \times \dfrac{\sqrt{2}}{2} = 5\sqrt{2}
Question 5: [2 marks]
ABC is a right-angled triangle.
\angle ABC = 30 \degreeCB = 12 cm
Find the exact value of x, and choose the correct answer from below.
Answer type: Multiple choice type 1
A: 4\sqrt{3} cm
B: 6\sqrt{3} cm
C: 6
D: 6\sqrt{2}
ANSWER: A: 4\sqrt{3} cm
WORKING:
x = 12 \tan(30 \degree) = 12 \times \dfrac{\sqrt{3}}{3} = 4\sqrt{3}
Question 6: [4 marks]
Below is a parallelogram.
AC = x
AB = x+3
The area of the parallelogram is 20 \sqrt{3}.
Find the value of x.
Answer type: Simple text answer
ANSWER: 5 cm
WORKING:
Form a right angled triangle with a line coming down from C.
\text{Height } = x \times \sin(60) = \dfrac{x\sqrt{3}}{2}
\text{Area } = \text{base } \times \text{height } = (x+3) \times \dfrac{x\sqrt{3}}{2} = \dfrac{x^2\sqrt{3}}{2} + \dfrac{3x\sqrt{3}}{2}
\dfrac{x^2\sqrt{3}}{2} + \dfrac{3x\sqrt{3}}{2} = 20\sqrt{3}
Which can be rearranged to be
x^2+3x-40=0
Use the quadratic formula to find
x = \dfrac{-3 \pm \sqrt{169}}{2} = 5 \text{ or } -8
Since x is a physical length, it must be positive, so x = 5 cm.