At the foot of a mountain, the elevation of it summit is 45°; after ascending 1000 m towards the mountain up a slope of 30° inclination, the elevation is found to be 60°. Find the height of the mountain.

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#### Solution

Suppose, AB is a mountain of height *t + x. *

\[\text{ In } \bigtriangleup DFC, \]

\[\sin30° = \frac{x}{1000} \]

\[ \Rightarrow x = 1000 \times \left( \frac{1}{2} \right) = 500 m\]

\[\text{ and } \]

\[\tan30° = \frac{x}{y}\]

\[ \Rightarrow y = \frac{x}{\tan30°} = 500\sqrt{3}\]

\[\text{ In } ∆ ABC, \]

\[\tan45°= \frac{t + x}{y + z}\]

\[ \Rightarrow t + x = y + z . . . \left( 1 \right)\]

\[\text{ In } ∆ ADE, \]

\[\tan60° = \frac{t}{z}\]

\[ \Rightarrow t = \sqrt{3}z . . . \left( 2 \right)\]

\[\text{ From } \left( 1 \right) \text{ and } \left( 2 \right), \text{ we have } \]

\[\sqrt{3}z + x = y + z\]

\[ \Rightarrow z\left( \sqrt{3} - 1 \right) = 500\left( \sqrt{3} - 1 \right)\]

\[ \Rightarrow z = 500 m\]

\[ \therefore t = \sqrt{3}z = 500\sqrt{3}\]

Hence, height of the mountain =\[t + x = 500\sqrt{3} + 500 = 500\left( \sqrt{3} + 1 \right) m\]

\[\sin30° = \frac{x}{1000} \]

\[ \Rightarrow x = 1000 \times \left( \frac{1}{2} \right) = 500 m\]

\[\text{ and } \]

\[\tan30° = \frac{x}{y}\]

\[ \Rightarrow y = \frac{x}{\tan30°} = 500\sqrt{3}\]

\[\text{ In } ∆ ABC, \]

\[\tan45°= \frac{t + x}{y + z}\]

\[ \Rightarrow t + x = y + z . . . \left( 1 \right)\]

\[\text{ In } ∆ ADE, \]

\[\tan60° = \frac{t}{z}\]

\[ \Rightarrow t = \sqrt{3}z . . . \left( 2 \right)\]

\[\text{ From } \left( 1 \right) \text{ and } \left( 2 \right), \text{ we have } \]

\[\sqrt{3}z + x = y + z\]

\[ \Rightarrow z\left( \sqrt{3} - 1 \right) = 500\left( \sqrt{3} - 1 \right)\]

\[ \Rightarrow z = 500 m\]

\[ \therefore t = \sqrt{3}z = 500\sqrt{3}\]

Hence, height of the mountain =\[t + x = 500\sqrt{3} + 500 = 500\left( \sqrt{3} + 1 \right) m\]

Concept: Sine and Cosine Formulae and Their Applications

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