Question

. A TV tower stands vertically on a
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.
D
From the top of a 7 m high building, the angle of el​

Answer 1

Correct Question -

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 joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°.

To find :

Find the height of the tower and the width of the canal.

Solution :

See the attachment .

Let the height of the tower AB be x metres .

Let the width of the canal BC be w meters .

In ∆ ABC

Tan 60° = √3

=> x/w = √3

=> x = w √3 .

In ∆ ABD

x / ( w + 20 ) = 1/√3

=> √3 x = w + 20

=> √3 ( w √3 ) = w + 20

=> 3w = w + 20

=> 2w = 20

=> w = 10 m

x = w √3 = 10√3 m .

Thus , the height of the tower is 10√3 m and the width of the canal is 10 m .

Answer :

The height of the tower is 10√3 m and the width of the canal is 10 m .

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Answer 2

$\huge\bf{\underline{\underline{Appropriate\:Question:}}}$

A TV tower stands vertically on a 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 joining 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.

$\huge\bf{\underline{\underline{Solution:}}}$

Suppose the tower be line AB.

therefore, height of the tower = AB

It is given that, from the point on the other bank directly opposite the tower, angle of elevation of the top of the tower is 60°.

therefore,$\displaystyle{\sf{ \angle ACB = 60°}}$

Again, angle of the elevation from the point D to the top of the tower = 30°

therefore,$\displaystyle{\sf{ \angle ADB = 30°}}$

and DC = 20 m.

To Find : the height of the tower AB and the width of the canal BC.

because the tower is perpendicular to the ground.

$\displaystyle{\sf{ \angle ABD = 90°}}$

In right angled triangle ACB,

$\displaystyle{\sf{tan\:C = \dfrac{side\:opposite\:to\:angle\:C}{side\:adjacent\:to\:angle\:C}}}$

:$\Rightarrow{\sf{tan\:60°= \dfrac{AB}{BC}}}$

:$\Rightarrow{\sf{ \sqrt{3} = \dfrac{AB}{BC}}}$

:$\Rightarrow{\sf{ \sqrt{3}BC = AB\:\:\:\:...(1)}}$

In right triangle ADB,

$\displaystyle{\sf{tan\:D = \dfrac{side\:opposite\:to\:angle\:D}{side\:adjacent\:to\:angle\:D}}}$

:$\Rightarrow{\sf{tan\:30°= \dfrac{AB}{BD}}}$

:$\Rightarrow{\sf{ \dfrac{1}{ \sqrt{3}}= \dfrac{AB}{BD}}}$

:$\Rightarrow{\sf{ \dfrac{1}{ \sqrt{3}}= \dfrac{AB}{BD}}}$

:$\Rightarrow{\sf{ \dfrac{BD}{ \sqrt{3}}=AB\:\:\:...(2)}}$

From (1) and (2),we get

:$\Rightarrow{\sf{ \sqrt{3}BC = \dfrac{BD}{ \sqrt{3}}}}$

:$\Rightarrow{\sf{ \sqrt{3}BC \times \sqrt{3}= BD}}$

:$\Rightarrow{\sf{ 3BC = BD}}$

:$\Rightarrow{\sf{ 3BC = BC+CD}}$

:$\Rightarrow{\sf{ 3BC = BC+20}}$

:$\Rightarrow{\sf{ 3BC - BC= 20}}$

:$\Rightarrow{\sf{ 2BC= 20}}$

:$\Rightarrow{\sf{ BC= \dfrac{20}{2}}}$

:$\Rightarrow{\sf{ BC= 10}}$

$\large{\boxed{\sf{\red{Hence,\:the\:width\:of\:the\:canal\:is\:10m}}}}$

From (1),

:$\Rightarrow{\sf{ \sqrt{3}BC = AB}}$

:$\Rightarrow{\sf{ \sqrt{3}10 = AB}}$

:$\Rightarrow{\sf{AB=10 \sqrt{3}}}$

$\large{\boxed{\sf{\red{Hence,\:the\:height\:of\:the\:tower\:is\:10 \sqrt{3}m}}}}$

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