Question

question fifth and sixth

Answer 1

Q5 let AB and CD be the Tower of height X and BD be the road of 80m where point P lies. let BD = x then PD=80-x
then in rt triangle ABP
tan 60 =AB/BP
√3=AB/x
x√3=AB. eq 1
similarly in right triangle PCD
Tan 30=CD/DP
1/√3=CD/80-x
80-x/√3=AB. eq 2
from eq 1 and eq 2
x√3=80-x/√3
3x=80-x
4x=80
x=20
BP=x=20m
PD=80-x=60m
ht of pole =AB=x√3=20√3m