(D) Let the height of the eagle be $h = 100 \ m$. Let the initial horizontal distance from the prey be $d_1$ and the final horizontal distance be $d_2

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(D) Let the height of the eagle be $h = 100 \ m$. Let the initial horizontal distance from the prey be $d_1$ and the final horizontal distance be $d_2$.From the first position,$\tan(30^\circ) = \frac{h}{d_1}$ $\Rightarrow \frac{1}{\sqrt{3}} = \frac{100}{d_1}$ $\Rightarrow d_1 = 100\sqrt{3} \ m$.From the second position,$\tan(45^\circ) = \frac{h}{d_2}$ $\Rightarrow 1 = \frac{100}{d_2}$ $\Rightarrow d_2 = 100 \ m$.The distance $x$ covered is $d_1 - d_2 = 100\sqrt{3} - 100 \ m$.

Class 11 Mathematics Trigonometrical Equations Height and Distance

An eagle flying $100 \ m$ above the ground looks at its prey on the ground at an angle of depression of $30^\circ$. After flying a distance of $'x' \ m$ towards the prey at the same height,the angle of depression becomes $45^\circ$. The value of $x$ will be:

A
$100\sqrt{5}$
B
$100\sqrt{3}+100$
C
$100$
D
$100\sqrt{3}-100$

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An eagle flying $100 \ m$ above the ground looks at its prey on the ground at an angle of depression of $30^\circ$. After flying a distance of $'x' \ m$ towards the prey at the same height,the angle of depression becomes $45^\circ$. The value of $x$ will be:

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At a point on the ground,the angle of elevation of a tower is such that its cotangent is $3/5$. On walking $32 \, m$ towards the tower,the cotangent of the angle of elevation becomes $2/5$. The height of the tower is .... $m$

The angles of elevation of the top of a tower $A$ from the top $B$ and bottom $D$ of a building of height $a$ are $30^\circ$ and $45^\circ$ respectively. If the tower and the building stand at the same level,then the height of the tower is

$AB$ is a vertical pole resting at the end $A$ on the level ground. $P$ is a point on the level ground such that $AP = 3 \, AB$. If $C$ is the mid-point of $AB$ and $CB$ subtends an angle $\beta$ at $P$,the value of $\tan \beta$ is

$A$ horizontal park is in the shape of a triangle $OAB$ with $AB = 16$. $A$ vertical lamp post $OP$ is erected at the point $O$ such that $\angle PAO = \angle PBO = 15^{\circ}$ and $\angle PCO = 45^{\circ}$,where $C$ is the midpoint of $AB$. Then $(OP)^{2}$ is equal to.

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