A bird is sitting on the top of a vertical pole $$20$$ m high and its elevation from a point $$O$$ on the ground is $${45^ \circ }$$. It files off horizontally straight away from the point $$O$$. After one second, the elevation of the bird from $$O$$ is reduced to $${30^ \circ }$$. Then the speed (in m/s) of the bird is
A
$$20\sqrt 2 $$
B
$$20\left( {\sqrt 3 - 1} \right)$$
C
$$40\left( {\sqrt 2 - 1} \right)$$
D
$$40\left( {\sqrt 3 - \sqrt 2 } \right)$$
Explanation
Let the speed be $$y$$ $$m/sec$$.
Let $$AC$$ be the vertical pole of height $$20$$ $$m.$$
Let $$O$$ be the point on the ground such that $$\angle AOC = {45^ \circ }$$
Let $$OC = x$$
Time $$t=1$$ $$s$$
From $$\Delta AOC,\,\,\tan {45^ \circ } = {{20} \over x}\,\,\,\,\,\,\,.....\left( i \right)$$
and from $$\Delta BOD,\,\,\tan {30^ \circ } = {{20} \over {x + y}}...\left( {ii} \right)$$
$$AB$$ is a vertical pole with $$B$$ at the ground level and $$A$$ at the top. $$A$$ man finds that the angle of elevation of the point $$A$$ from a certain point $$C$$ on the ground is $${60^ \circ }$$. He moves away from the pole along the line $$BC$$ to a point $$D$$ such that $$CD=7$$ m. From $$D$$ the angle of elevation of the point $$A$$ is $${45^ \circ }$$. Then the height of the pole is
A tower stands at the centre of a circular park. $$A$$ and $$B$$ are two points on the boundary of the park such that $$AB(=a)$$ subtends an angle of $${60^ \circ }$$ at the foot of the tower, and the angle of elevation of the top of the tower from $$A$$ or $$B$$ is $${30^ \circ }$$. The height of the tower is
A
$$a/\sqrt 3 $$
B
$$a\sqrt 3 $$
C
$$2a/\sqrt 3 $$
D
$$2a\sqrt 3 $$
Explanation
In the $$\Delta AOB,\,\,\angle AOB = {60^ \circ },$$ and
$$\angle OBA = \angle OAB$$
(since $$OA=OB=AB$$ radius of same circle).
$$\therefore$$ $$\Delta AOB$$ is a equilateral triangle.
Let the height of tower is $$h$$
$$m.$$ Given distance between two points $$A$$ & $$B$$ lie on boundary of
circular park, subtends an angle of $${60^ \circ }$$ at the foot of the tower
is $$AB$$ i.e. $$AB$$$$=a.$$ A tower $$OC$$ stands at the center of a circular
park. Angle of elevation of the top of the tower from $$A$$ and $$B$$ is $${30^ \circ }$$ .
In $$\Delta OAC\,\,\tan {30^ \circ } = {h \over a}$$