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Some Applications Of Trigonometry
EXERCISE 9.1
CBSE_NCERT_TEXTBOOK
Grade 10
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 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.
Step-by-Step Solution
Key Concept: Use the definition of tangent in a right‑angled triangle: \(\tan\theta = \frac{\text{opposite}}{\text{adjacent}}\). Let the width of the canal be \(w\) metres and the height of the tower be \(h\) metres. Write two equations using the given angles of elevation and the two distances from the foot of the tower, then solve the simultaneous equations.
1. Introduce variables\
Let
- \(w\) = width of the canal (distance from the point A on the opposite bank to the foot of the tower B).\
- \(h\) = height of the TV tower.
2. First observation (point A)\
From point A the angle of elevation to the top of the tower is \(60^{\circ}\).\
In the right‑angled triangle \(\triangle ABT\) (where T is the top of the tower),\
\[\tan 60^{\circ}=\frac{h}{w}\]
Since \(\tan 60^{\circ}=\sqrt{3}\), we get\
\[\sqrt{3}=\frac{h}{w}\quad\Rightarrow\quad h=\sqrt{3}\,w \tag{1}\]
3. Second observation (point C)\
Point C is 20 m away from A away from the tower along the same straight line (as shown in Fig. 9.12). Hence the distance from C to the foot of the tower is \(w+20\) metres.
The angle of elevation from C is \(30^{\circ}\).\
Using \(\tan 30^{\circ}=\frac{1}{\sqrt{3}}\),\
\[\tan 30^{\circ}=\frac{h}{w+20}\]
gives\
\[\frac{1}{\sqrt{3}}=\frac{h}{w+20}\quad\Rightarrow\quad h=\frac{w+20}{\sqrt{3}} \tag{2}\]
4. Equate the two expressions for \(h\)\
From (1) and (2):\
\[\sqrt{3}\,w = \frac{w+20}{\sqrt{3}}\]
Multiply both sides by \(\sqrt{3}\):\
\[3w = w + 20\]
\[2w = 20\]
\[w = 10\ \text{metres}\]
5. Find the height \(h\)\
Substitute \(w = 10\) into (1):\
\[h = \sqrt{3}\times 10 = 10\sqrt{3}\ \text{metres}\]
Numerically, \(h \approx 17.3\) m.
6. Answer\
- Width of the canal = \(10\) m\
- Height of the TV tower = \(10\sqrt{3}\) m (≈ 17.3 m).
Let
- \(w\) = width of the canal (distance from the point A on the opposite bank to the foot of the tower B).\
- \(h\) = height of the TV tower.
2. First observation (point A)\
From point A the angle of elevation to the top of the tower is \(60^{\circ}\).\
In the right‑angled triangle \(\triangle ABT\) (where T is the top of the tower),\
\[\tan 60^{\circ}=\frac{h}{w}\]
Since \(\tan 60^{\circ}=\sqrt{3}\), we get\
\[\sqrt{3}=\frac{h}{w}\quad\Rightarrow\quad h=\sqrt{3}\,w \tag{1}\]
3. Second observation (point C)\
Point C is 20 m away from A away from the tower along the same straight line (as shown in Fig. 9.12). Hence the distance from C to the foot of the tower is \(w+20\) metres.
The angle of elevation from C is \(30^{\circ}\).\
Using \(\tan 30^{\circ}=\frac{1}{\sqrt{3}}\),\
\[\tan 30^{\circ}=\frac{h}{w+20}\]
gives\
\[\frac{1}{\sqrt{3}}=\frac{h}{w+20}\quad\Rightarrow\quad h=\frac{w+20}{\sqrt{3}} \tag{2}\]
4. Equate the two expressions for \(h\)\
From (1) and (2):\
\[\sqrt{3}\,w = \frac{w+20}{\sqrt{3}}\]
Multiply both sides by \(\sqrt{3}\):\
\[3w = w + 20\]
\[2w = 20\]
\[w = 10\ \text{metres}\]
5. Find the height \(h\)\
Substitute \(w = 10\) into (1):\
\[h = \sqrt{3}\times 10 = 10\sqrt{3}\ \text{metres}\]
Numerically, \(h \approx 17.3\) m.
6. Answer\
- Width of the canal = \(10\) m\
- Height of the TV tower = \(10\sqrt{3}\) m (≈ 17.3 m).
Correct Answer: Width of the canal = 10 m; Height of the tower = 10√3 m (≈ 17.3 m).
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