Question

In triangle ABC , AB =10cm ,AC=8cm ,angle A =50 (a) find the perpendicular distance from C to AB
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Answer 1

Answer:

Let the perpendicular be CD on AB.

In triangle ADC

tan60=CD/AD = 3^1/2

CD = AD*3^1/2 ………………..(1)

tan45 = CD/BD = 1

CD = BD ………………(2)

From (1) and (2)

we get AD*3^1/2 = BD

and AD + BD = 10cm

so AD(3^1/2 + 1) = 10cm

AD = 5(3^1/2 -1)

AD = 5*0.732 = 3.66cm

CD = AD*3^1/2

CD = 3.66*1.732 = 6.33912cm