(A) Let the height of the lighthouse be $AB = 100 \, m$. Let the two boats be at points $C$ and $D$ such that $C$ is closer to the lighthouse.In $\tri

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(A) Let the height of the lighthouse be $AB = 100 \, m$. Let the two boats be at points $C$ and $D$ such that $C$ is closer to the lighthouse.In $\triangle ABC$,$\tan(45^{\circ}) = \frac{AB}{BC} \implies 1 = \frac{100}{BC} \implies BC = 100 \, m$.In $\triangle ABD$,$\tan(30^{\circ}) = \frac{AB}{BD} \implies \frac{1}{\sqrt{3}} = \frac{100}{BD} \implies BD = 100\sqrt{3} \, m$.Using $\sqrt{3} \approx 1.732$,$BD = 100 \times 1.732 = 173.2 \, m$.The distance between the two boats is $CD = BD - BC = 173.2 - 100 = 73.2 \, m$.

Class 10 Mathematics Some Applications of Trigonometry Mix Examples - Some Applications of Trigonometry

Medium

Two boats are anchored in the same direction from a $100 \, m$ high lighthouse. The angles of depression of the boats from the top of the lighthouse are $30^{\circ}$ and $45^{\circ}$. Find the distance between these two boats (in $m$).

A
$73.2$
B
$61.5$
C
$55.4$
D
$49.1$

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Some Applications of Trigonometry

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Mix Examples - Some Applications of Trigonometry

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Two boats are anchored in the same direction from a $100 \, m$ high lighthouse. The angles of depression of the boats from the top of the lighthouse are $30^{\circ}$ and $45^{\circ}$. Find the distance between these two boats (in $m$).

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Watching from the top of a $150\, m$ high tower,the angles of depression of two houses situated in the east and west of the tower are found to be $30^{\circ}$ and $45^{\circ}$ respectively. Find the distance between the houses in meters.

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