Introduction to trigonometry 105
- Evaluate correct to 5 significant figures:
(a) cosec(− 143 ◦) (b) cot(− 252 ◦)
(c) sec(− 67 ◦[ 22 ′)
(a)− 1. 6616 (b)− 0. 32492
(c) 2. 5985
]
11.5 Solution of right-angled triangles
To ‘solve a right-angled triangle’ means ‘to find the
unknown sides and angles’. This is achieved by using
(i) the theorem of Pythagoras, and/or (ii) trigonometric
ratios. This is demonstrated in the following problems.
Problem 21. In trianglePQRshown in
Fig. 11.14, find the lengths ofPQandPR.
Q R
P
7.5 cm
388
Figure 11.14
tan38◦=
PQ
QR
=
PQ
7. 5
hence PQ=^7 .5tan38◦=^7.^5 (^0.^7813 )
=5.860cm
cos38◦=
QR
PR
=
7. 5
PR
hence PR=
7. 5
cos38◦
=
7. 5
0. 7880
=9.518cm
[Check: Using Pythagoras’ theorem
( 7. 5 )^2 +( 5. 860 )^2 = 90. 59 =( 9. 518 )^2 ]
Problem 22. Solve the triangleABCshown in
Fig. 11.15.
A
C
B
37 mm
35 mm
Figure 11.15
To ‘solve triangleABC’ means ‘to find the length
ACand anglesBandC’
sinC=
35
37
= 0. 94595
hence∠C=sin−^10. 94595 = 71. 08 ◦= 71 ◦ 5 ′.
∠B= 180 ◦− 90 ◦− 71 ◦ 5 ′= 18 ◦ 55 ′ (since angles in a
triangle add up to 180◦)
sinB=
AC
37
hence AC=37sin18◦ 55 ′= 37 ( 0. 3242 )
=12.0mm
or, using Pythagoras’ theorem, 37^2 = 352 +AC^2 , from
which,AC=
√
( 372 − 352 )=12.0mm.
Problem 23. Solve triangleXYZgiven
∠X= 90 ◦,∠Y= 23 ◦ 17 ′andYZ= 20 .0mm.
Determine also its area.
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
Fig. 11.16.
∠Z= 180 ◦− 90 ◦− 23 ◦ 17 ′= 66 ◦ 43 ′
sin23◦ 17 ′=
XZ
20. 0
X Y
Z
20.0mm
238179
Figure 11.16
hence XZ= 20 .0sin23◦ 17 ′
= 20. 0 ( 0. 3953 )=7.906mm
cos23◦ 17 ′=
XY
20. 0
hence XY= 20 .0cos23◦ 17 ′
= 20. 0 ( 0. 9186 )=18.37mm
[Check: Using Pythagoras’ theorem
(18.37)^2 +( 7. 906 )^2 = 400. 0 =( 20. 0 )^2 ]