Basic Engineering Mathematics, Fifth Edition

Introduction to trigonometry 189

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 21. Solve the triangleABCshown in

Figure 21.22

A

C

B

37mm

35mm

Figure 21.22

To ‘solve the triangleABC’ means ‘to find the length
AC and anglesBandC’.

sinC=

35

37

= 0. 94595 ,hence

C=sin−^10. 94595 = 71. 08 ◦

B= 180 ◦− 90 ◦− 71. 08 ◦= 18. 92 ◦(sincetheangles in
a triangle add up to 180◦)

sinB=

AC

37

,hence

AC=37sin18. 92 ◦= 37 ( 0. 3242 )=12.0mm

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

whichAC=

√

( 372 − 352 )=12.0mm.

Problem 22. Solve triangleXY Zgiven

∠X= 90 ◦,∠Y= 23 ◦ 17 ′andYZ= 20 .0mm

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

20.0mm

238179

Figure 21.23

sin23◦ 17 ′=

XZ

20. 0

,henceXZ= 20 .0sin23◦ 17 ′

= 20. 0 ( 0. 3953 )= 7 .906mm

cos23◦ 17 ′=

XY

20. 0

,henceXY= 20 .0cos23◦ 17 ′

= 20. 0 ( 0. 9186 )=18.37mm

Check: using Pythagoras’ theorem,

( 18. 37 )^2 +( 7. 906 )^2 = 400. 0 =( 20. 0 )^2 ,

Now try the following Practice Exercise

PracticeExercise 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)

17.0

708

228

x

13.0

(a)

(b)

x

15.0

Figure 21.24