Question

3. In triangle ABC AB = 8 cm Angle A = 50 degree Angle B = 70 degree
a] find the perpendicular distance from 'C' to AB

Answer 1

Answer:

Let CD be perpendicular from C to AB.

(a) In ΔACD,

sin50

o

=

AC

CD

⇒0.7660=

5

CD

⇒CD=0.7660×5=3.83cm

(b) cos50

o

=

AC

AD

⇒0.6428=

5

AD

⇒AD=0.6428×5=3.214

DB = AB - AD = (8 - 3.214) cm = 4.786 cm

In right triangle CDB,

BC

2

=CD

2

+DB

2

=(3.83)

2

+(4.786)

2

=14.6+22.91

=37.58

⇒B=

37.58

cm