Question

hellppp.....
class 10

Answer 1

Answer:

I'm not sure

Step-by-step explanation:

Let , the point is o and,

Height of Pillars = h

and Distance between AB = 150 m (here DA and CB are pillars)

So,

Ao = 150 - oB

in Triangle(DAo),

tanD(D = 60)= Ao / DA =>  √3 = (150- ob)/ h

so, h = (150 - oB) /√3

Similarly in Triangle(CBo)

tanC(C = 30) = oB / CB => 1/√3 = oB / h

so, h = √3oB

By Equating Both h We get,

(150 - oB) / √3 = √3oB

150-oB = √3×√3 oB = 3oB

150 = 3oB + oB = 4oB

oB = 37.5

h = 37.5√3