Basic Engineering Mathematics, Fifth Edition

Angles and triangles 177

508

708

c 5 12.0cm

B

a C

A

508

608

f 5 5.0cm

d 5 4.42cm

E F

D

Figure 20.39

In triangleABC, 50 ◦+ 70 ◦+∠C= 180 ◦, from which
∠C= 60 ◦.
In triangle DEF,∠E= 180 ◦− 50 ◦− 60 ◦= 70 ◦.
Hence, trianglesABCandDEFare similar, since their
angles are the same. Since corresponding sides are in
proportion toeach other,

a

d

=

c

f

i.e.

a

4. 42

=

12. 0

5. 0

Hence,side,a=

12. 0

5. 0

( 4. 42 )= 10 .61cm.

Problem 27. In Figure 20.40, find the dimensions

markedrandp

558

358

x 5 7.44cm

q 5 6.82cm z

5

12.97cm

y^5 10.63cm

Y

P Q

R

Z

X

p

r

Figure 20.40

In trianglePQR,∠Q= 180 ◦− 90 ◦− 35 ◦= 55 ◦.
In triangleXYZ,∠X= 180 ◦− 90 ◦− 55 ◦= 35 ◦.
Hence, trianglesPQRandZYXare similar since their
angles are the same. The triangles may he redrawn as
shown in Figure 20.41.

358

558

q 5 6.82cm y 5 10.63cm

x^5

7.44cm

z 5

12.97cm

Z X

Y

358

558

PR

r p

Q

Figure 20.41

By proportion:

p

z

=

r

x

=

q

y

i.e.

p

12. 97

=

r

7. 44

=

6. 82

10. 63

from which, r= 7. 44

(

6. 82

10. 63

)

= 4 .77cm

By proportion:

p

z

=

q

y

i.e.

p

12. 97

=

6. 82

10. 63

Hence, p= 12. 97

(

6. 82

10. 63

)

= 8 .32cm

Problem 28. In Figure 20.42, show that triangles

CBDandCAEare similar and hence find the length

ofCDandBD

B

10

6

9

12

C

D

E

A

Figure 20.42

SinceBDis parallel toAEthen∠CBD=∠CAEand

∠CDB=∠CEA(corresponding angles between paral-

lel lines). Also,∠Cis common to trianglesCBDand

CAE.

Since the angles in triangleCBDare the same as in

triangleCAE, the triangles are similar. Hence,

by proportion:

CB

CA

=

CD

CE

(

=

BD

AE

)

i.e.

9

6 + 9

=

CD

12

,from which

CD= 12

(

9

15

)

= 7 .2cm

Also,

9

15

=

BD

10

,from which

BD= 10

(

9

15

)

=6cm