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

## 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 \(BC=10 \,\rm{m}\)