Basic Engineering Mathematics, Fifth Edition

188 Basic Engineering Mathematics

13. Evaluate, correct to 4 decimal places,
    4 .5cos67◦ 34 ′−sin90◦
    2tan45◦


14. Evaluate, correct to 4 significant figures,
    (3sin37. 83 ◦)( 2 .5tan57. 48 ◦)
    4 .1cos12. 56 ◦

21.5 Solving right-angled triangles

‘Solving a right-angled triangle’ means ‘finding the

unknown sides and angles’. This is achieved using

(a) the theorem of Pythagoras and/or

(b) trigonometric ratios.

Six pieces of information describeatrianglecompletely;

i.e., three sides and three angles. As long as at least

three facts are known, the other three can usually be

calculated.

Here are some worked problems to demonstrate the

solution of right-angled triangles.

Problem 18. In the triangleABCshown in

Figure 21.19, find the lengthsACandAB

A

B C

428

6.2mm

Figure 21.19

There is usually more than one way to solve such a

triangle.

In triangleABC,

tan42◦=

AC

BC

=

AC

6. 2

(Remember SOH CAH TOA)

Transposing gives

AC= 6 .2tan42◦=5.583mm

cos42◦=

BC

AB

=

6. 2

AB

,from which

AB=

6. 2

cos42◦

=8.343mm

Alternatively, by Pythagoras, AB^2 =AC^2 +BC^2

from which AB=

√

AC^2 +BC^2 =

√

5. 5832 + 6. 22

=

√

69. 609889 =8.343mm.

Problem 19. Sketch a right-angled triangleABC

such thatB= 90 ◦,AB=5cmandBC=12cm.

Determine the length ofACand hence evaluate

sinA,cosCand tanA

TriangleABCis shown in Figure 21.20.

A

B C

5cm

12cm

Figure 21.20

By Pythagoras’ theorem, AC=

√

52 + 122 = 13

By definition: sinA=

opposite side

hypotenuse

=

12

13

or0.9231

(Remember SOH CAH TOA)

cosC=

adjacent side

hypotenuse

=

12

13

or0.9231

and tanA=

opposite side

adjacent side

=

12

5

or2.400

Problem 20. In trianglePQRshown in

Figure 21.21, find the lengths ofPQandPR

Q R

P

7.5cm

388

Figure 21.21

tan38◦=

PQ

QR

=

PQ

7. 5

,hence

PQ= 7 .5tan38◦= 7. 5 ( 0. 7813 )=5.860cm