Higher Engineering Mathematics, Sixth Edition

Introduction to trigonometry 105

27. 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 ]