Introduction to trigonometry 189
cos 38
◦
=
QR
PR
=
7.5
PR
, hence
PR =
7.5
cos 38 ◦ =
7.5
0.7880
= 9.518 cm
Check: using Pythagoras’ theorem,
(7.5) 2 + (5.860) 2 = 90.59 = (9.518) 2 .
Problem 21. Solve the triangle ABC shown in
Figure 21.22
A
C
B
37 mm
35 mm
Figure 21.22
To ‘solve the triangle ABC’ means ‘to find the length
AC and angles B and C’.
sin C =
35
37
= 0.94595, hence
C = sin
−1 0.94595 = 71.08
◦
B = 180 ◦ − 90 ◦ − 71.08 ◦ = 18.92 ◦ (since the angles in
a triangle add up to 180 ◦ )
sin B =
AC
37
, hence
AC = 37 sin 18.92
◦
= 37(0.3242) = 12.0 mm
or, using Pythagoras’ theorem, 37 2 = 35 2 + AC 2 , from
which AC =
(37 2 − 35 2 ) = 12.0 mm.
Problem 22. Solve triangle XYZ given
∠X = 90 ◦ , ∠Y = 23 ◦ 17 and YZ = 20.0 mm
It is always advisable to make a reasonably accurate
sketch so as to visualize the expected magnitudes of
unknown sides and angles. Such a sketch is shown in
Figure 21.23.
∠Z = 180
◦
− 90
◦
− 23
◦ 17
= 66
◦ 43
X
Y
Z
20.0 mm
238179
Figure 21.23
sin 23
◦ 17
=
XZ
20.0
, hence XZ = 20.0 sin 23
◦ 17
= 20.0(0.3953) = 7.906 mm
cos 23
◦ 17
=
XY
20.0
, hence XY = 20.0 cos 23
◦ 17
= 20.0(0.9186) = 18.37 mm
Check: using Pythagoras’ theorem,
(18.37) 2 + (7.906) 2 = 400.0 = (20.0) 2 ,
Now try the following Practice Exercise
Practice Exercise 85 Solving right-angled
triangles (answers on page 349)
1. Calculate the dimensions shown as x in
Figures 21.24(a) to (f), each correct to 4
significant figures.
298
x
(c)
1
7
.
0
708
228
x
13.0
(a)
(b)
x
15.0
Figure 21.24