11. A TV tower stands vertically on a bank
of a canal. From a point on the other
bank directly opposite the tower, the
angle of elevation of the top of the
tower is 60°. From another point 20 m
away from this point on the line joing
this point to the foot of the tower, the
angle of elevation of the top of the
tower is 30° (see Fig. 9.12). Find the
height of the tower and the width of
the canal.
of a canal. From a point on the other
bank directly opposite the tower, the
angle of elevation of the top of the
tower is 60°. From another point 20 m
away from this point on the line joing
this point to the foot of the tower, the
angle of elevation of the top of the
tower is 30° (see Fig. 9.12). Find the
height of the tower and the width of
the canal.
Add an Answer
More Questions From This User See All
Answers & Comments
━━━━━━━━━━━━━━━━━━━━━━━━
According to figure
CD = 20m
∠ACB = 60°
∠ADB = 30°
◆ ▬▬▬▬▬▬ ❴✪❵ ▬▬▬▬▬▬ ◆
We have to find the value of AB and BD
↪In ∆ABC,
tan(theta)= p/b
where p = perpendicular and b = base of triangle
AB/BC
√3=AB/BC
AB=√3BC.........................(i)
━━━━━━━━━━━━━━━━━━━━━━━━
↪In ∆ABD,
tan(30°)= AB/BC+20
AB = BC+20/√3................(ii)
from equation (i) & (ii)
√3BC=BC+20/√3
3BC=BC+20
BC = 10m
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
↪Then,Putting the value of BC in (i) we get;
AB = 10√3 m (which is the height of the tower)
❄️Width of canal = 10 m❄️
=========================================