CHAPTER 17. TRIGONOMETRY 17.5
- W XY Z is a trapezium (W X� XZ) with W X = 3 m;
Y Z = 1,5 m;Zˆ = 120◦andWˆ = 30◦.
Determine the distances XZ and XY. 1 ,5 m
3 m
30 ◦
120 ◦
W
XY
Z
- On a flight from Johannesburg to Cape Town, the pilot discovers that he has been
flying 3 ◦off course. At this pointthe plane is 500 km from Johannesburg.The direct
distance between CapeTown and Johannesburgairports is 1 552 km. Determine, to
the nearest km:
(a) The distance the plane has to travel to get to Cape Town and hence the extra
distance that the plane has had to travel due to the pilot’s error.
(b) The correction, to one hundredth of a degree, to the plane’s heading (or direc-
tion).
- ABCD is a trapezium (i.e. AB� CD). AB = x;
BADˆ = a; BCDˆ = b and BDCˆ = c.
Find an expression for the length of CD in terms of
x, a, b and c.
A B
D C
a
c b
x
- A surveyor is tryingto determine the distance between
points X and Z. However the distancecannot be de-
termined directly as a ridge lies between the two points.
From a point Y which is equidistant from X and Z, he
measures the angle XY Zˆ.
(a) If XY = x and XY Zˆ = θ, show that XZ =
x
�
2(1− cos θ).
(b) Calculate XZ (to the nearest kilometre) if x =
240 km and θ = 132◦.
Y
X Z
x θ
- Find the area of W XY Z (to two decimal places):
W
X
Z Y
120 ◦ 3
4
3 , 5
- Find the area of theshaded triangle in termsof x, α, β,
θ and φ:
A B
E D C
x
θ β α
φ
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(1.) 0157 (2.) 0158 (3.) 0159 (4.) 015a (5.) 015b (6.) 015c
(7.) 015d
