A vertical pole fixed to the horizontal ground is divided in the ratio \(3 : 7\) by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a point on the ground \(18\text{ m}\) away from the base of the pole, then the height of the pole (in meters) is :
Show Hint
When parts of a vertical object subtend angles, remember that the "upper" part subtends an angle that is the difference between the angle subtended by the whole object and the angle subtended by the lower part.
Step 1: Understanding the Concept:
This problem uses basic trigonometry and the tangent double angle formula. We define the heights of the parts based on the ratio and equate the angles using the tangent function. Step 2: Detailed Explanation:
1. Let the total height of the pole be \(H\). The mark divides it in ratio \(3:7\).
Lower part \(h_1 = 0.3H\), Upper part \(h_2 = 0.7H\).
2. Let the point on ground be \(P\), at distance \(d = 18\text{m}\).
3. Let the lower part subtend angle \(\theta\) at \(P\). Then \(\tan \theta = \frac{0.3H}{18}\).
4. The upper part subtends the same angle \(\theta\). Thus, the whole pole subtends angle \(2\theta\) at \(P\).
5. \(\tan 2\theta = \frac{H}{18}\).
6. Use the formula \(\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}\):
\[ \frac{H}{18} = \frac{2 \cdot \frac{0.3H}{18}}{1 - (\frac{0.3H}{18})^2} \]
Divide by \(H/18\) (since \(H \neq 0\)):
\[ 1 = \frac{0.6}{1 - \frac{0.09 H^2}{324}} \implies 1 - \frac{0.09 H^2}{324} = 0.6 \]
\[ \frac{0.09 H^2}{324} = 0.4 \implies H^2 = \frac{0.4 \cdot 324}{0.09} = \frac{4 \cdot 324}{0.9} = \frac{40 \cdot 324}{9} \]
\[ H^2 = 40 \cdot 36 = 1440 \implies H = \sqrt{144 \cdot 10} = 12\sqrt{10} \] Step 3: Final Answer:
The height of the pole is \(12\sqrt{10}\text{ m}\).
