Ex.9.1 Q11 Some Applications of Trigonometry Solution - NCERT Maths Class 10

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Question

A TV tower stands vertically on the 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 \(60^\circ\). From another point \(20 \,\rm{m}\) 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 \(30^\circ\) see Figure. Find the height of the tower and the width of the canal.

Text Solution

   

What is Known?

(i) The angle of elevation of the top of the tower from a point on the other bank directly opposite the tower is \(60^\circ\)

(ii) From another point \(20\,\rm{ m}\) away from this point in (i) on the line joining this pointto the foot of the tower, the angle of elevation of the top of the tower is \(30^\circ\)

(iii)  \( CD =20\,\rm{ m}\)

What is Unknown?

Height of the tower \(=AB\)  and the width of canal \(=BC\)

Reasoning:

Trigonometric ratio involving \(CD,\, BC,\) angles and height of tower \(AB\) is \(tan\,\theta \).

Steps:

Considering \(\Delta ABC,\)

\[\begin{align} \tan 60 ^ { \circ } & = \frac { A B } { B C } \\ \sqrt { 3 } & = \frac { A B } { B C } \\ A B & = \sqrt { 3 } B C \ldots . ( 1 ) \end{align}\]

Considering \(\Delta ABD,\)

\[\begin{align}\text{tan 3}{{\text{0}}^{0}}&=\frac{AB}{BD} \\ \text{tan 3}{{\text{0}}^{0}}&=\frac{AB}{CD+BC} \\ \frac{1}{\sqrt{3}}&=\frac{BC\sqrt{3}\,}{20+BC}\qquad{From}\,(1) \\ 20+BC&=BC\sqrt{3}\times \sqrt{3} \\ 20+BC&=3BC \\ 3BC-BC&=20 \\ 2BC&=20 \\ BC&=10\end{align}\]

Substituting\( BC = 10 \,\rm{m}\) in Equation (\(1\)), we get

\(\text{AB = 10}\sqrt{\text{3}}\,\rm{m}\)

Height of the tower \(\text{AB = 10}\sqrt{\text{3}}\)

Width of the canal \(BC=10 \,\rm{m}\)