Question

plzz solve Q 20 in the following image <br /><br />

Answer 1

let BD = x ...and CD = 10-x
from triangle ABD,
height/base = tan 60
ie AD/BD=root 3
or AD/x=root 3
or AD= x root 3..... (1)
Again from triangle ACD
AD/DC=tan45
or AD= DCtan 45
or AD= (10-x)(1)= 10-x...... (2)
so from (1) and (2)....
x root 3= 10-x
or x (root3+1)=10
or x =10/(root 3 +1)
AD= (10 root3)/(root3+1)
simplifying it we get
AD=5root3 (root3-1)=5(3-root3)

Answer 2

Let BD is x , then DC = (10 - x)

now, use tan∅ for both traingles.

for ∆ABD,
tan60° = AD/BD
√3 = AD/x
AD = x√3 ______________(1)


for ∆ADC,
tan45° = AD/DC
1 = AD/(10 - x)
AD = 10 - x _____________(2)

from equations (1) and (2)

x√3 = 10 - x
x(√3 + 1) = 10
x = 10/(√3 + 1) = 5(√3 - 1)

now, AD = x√3
= 5(√3 -1)√3
= 5(3 - √3) cm