Question

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.​

Answer 1

$\huge{\underline{\underline{\rm{SolutiOn:-}}}}$

${\bf{\red{Given\:that}}}$

• Angle of elevation of top of toweris 60°&30°
• Distance between these angles is 20m.

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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

$\tan(60 {}^{} ) =$

AB/BC

√3=AB/BC

AB=√3BC.........................(i)

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↪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

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↪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❄️

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