Higher Engineering Mathematics

118 GEOMETRY AND TRIGONOMETRY

Figure 12.7

3. For the right-angled triangle shown in
   Fig. 12.8, find:

(a) sinα (b) cos[θ (c) tanθ

(a)

15

17

(b)

15

17

(c)

8

15

]

Figure 12.8

4. PointPlies at co-ordinate (−3, 1) and point
   Qat (5,−4). Determine
   (a) the distancePQ
   (b) the gradient of the straight linePQand
   (c) the anglePQmakes with the horizontal
   [(a) 9.434 (b)− 0 .625 (c) 32◦]

12.4 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) trigono-
metric ratios. This is demonstrated in the following
problems.

Problem 6. In trianglePQRshown in Fig. 12.9,

find the lengths ofPQandPR.

Figure 12.9

tan 38◦=

PQ

QR

=

PQ

7. 5

hence PQ= 7 .5 tan 38◦= 7 .5(0.7813)

=5.860 cm

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 7. Solve the triangleABCshown in

Fig. 12.10.

Figure 12.10

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=37 sin 18◦ 55 ′=37(0.3242)

=12.0 mm

or, using Pythagoras’ theorem, 37^2 = 352 +AC^2 ,

from which,AC=

√

(37^2 − 352 )=12.0 mm.

Problem 8. Solve triangle XYZ given

∠X= 90 ◦,∠Y= 23 ◦ 17 ′andYZ= 20 .0 mm.

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. 12.11.

∠Z= 180 ◦− 90 ◦− 23 ◦ 17 ′= 66 ◦ 43 ′

sin 23◦ 17 ′=

XZ

20. 0