Applications in Trigonometry Revision Ex 16 -17 (Sol)

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Tak Sun Secondary School
Marks
Form 3 Mathematics (Applications in Trigonometry)
Revision Exercise 2016-2017
Name : ___________________________
Class :
/35
_________ No. :_______
Section A : Multiple Choices (2 marks each, 10 marks in Total)
Choose the most suitable answer and write down into the corresponding boxes.
Time Allowed: 35 minutes
1. B
1.
2. C
3. A
4. A
Caroline rides a bicycle and moves up a road with gradient
5. A
1
. If her horizontal run is 2.4 km,
12
find her vertical rise.
A.
B.
C.
D.
2.
3.
0.05 km
0.2 km
5 km
12 km
If the gradient of a road is 0.23, what is the inclination of the road?
(Give the answer correct to 3 significant figures.)
A.
B.
12.0
12.9
C.
D.
13.0
13.9
The figure shows a contour map of scale 1 : 20 000. If the horizontal run of AB is 1 cm on the
map, find the inclination of AB, correct to the nearest degree.
A.
B.
C.
27
30
45
D.
60
1
4.
The angle of elevation of the top C of the building CD from the top of the building AB is 30°.
The angle of depression of the bottom D of the building CD from the top of the building AB is
50°. If the height of the building AB is 80 m, find the height of the building CD, correct to 3
significant figures.
5.
A.
B.
119 m
135 m
C.
D.
165 m
245 m
In the figure, E and F are two lifeguard towers 200 m apart along the shore. Given that the
bearings of the swimmer C from E and F are 045° and 315° respectively, find the shortest
distance from C to the shore.
A.
B.
C.
D.
100 m
150 m
200 m
300 m
Section B : (Short Questions) Total 25 marks
Please write all answers with steps in the spaces provided.
6.
Jane walks 12 m up a slope and her vertical rise is 5 m. Find the gradient of the stair, correct to 3
significant figures.
(4 marks)
In △ABC,
BC 2 + AC 2 = AB 2
AC 
1M
AB 2  BC 2
 12 2  5 2 m
 119 m
BC
Gradient of the stair 
AC
5m

119 m
 0.458 , cor. to 3 sig. fig.
1A
1M
1A
2
7.
The figure shows a contour map. Now a straight road is going to be constructed to connect E
and F. If the actual horizontal run between E and F is 420 m, find the gradient of the straight
road. Express your answer as a fraction.
100 m
80 m
60 m
40 m
F
20 m
0m
E
(5 marks)
1M+1A
Vertical rise of EF = (80 – 20) m = 60 m
Refer to the figure.
1M
FD
Gradient of the straight road 
ED
60 m

420 m
1

7
1M
1A
3
8. Cindy stands on the belvedere of the peak to see the pier on the opposite bank. Her eye level is
450 m above the horizontal. If the horizontal distance between Cindy and the pier is 5 400 m, find
the angle of depression of the pier from Cindy.
(Give the answer correct to 3 significant figures.)
(4 marks)
Refer to the figure.
Let the angle of depression required be θ.
1M
In △ABC,
∠ABC = 
450 m
AC

tan  
BC 5 400 m

1M
1M
 = 4.76, cor. to 3 sig. fig.

The angle of depression required is 4.76°.
4
1A
9.A boy rides a bicycle from position A to B in a direction of 056°, then he turns 74° left to C which
is due north of A. Find the whole circle bearing of B from C.
(5 marks)
With the notation in the figure,
In △ABC,
∠ACB + 56 = 74
∠ACB = 18
 = 180 – ∠ACB
1M
1A
1M
= 180 – 18
= 162
 The whole circle bearing of B from C is 162°.
1A
1A
5
10.Two speedboats A and B sail from the same position X at the same time. Speedboat A sails at a
speed of 40 km/h due south and speedboat B sails at a speed of 60 km/h due west. Find the reduced
bearing of A from B after 15 minutes, correct to the nearest degree.
N
B
60 km/h
X
40 km/h
A
(7 marks)
After 15 minutes,
distance between B and X = 60 
15
km
60
= 15 km
15
distance between A and X = 40 km
60
= 10 km
With the notation in the figure,
1A
1A
In △AXB,
∠AXB = 90
AX
tan  
BX
10 km

15 km
  = 33.69, cor. to 2 d.p.
 = 90 – 
1A
1M
= 90 – 33.69
= 56, cor. to the nearest degree
 The reduced bearing of A from B is S56°E.
1A
1A
1M
End of Paper
6
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