Height And Distance - Study Mode
[#41] The length of the shadow of a vertical pole on the ground is 36 m. If the angle of elevation of the sun at that time is $$ heta $$ such that $$sec heta = frac{{13}}{{12}},$$ xa0 then what is the height (in cm) of the pole?
Correct Answer
(D) 15
Explanation
Solution: [x08egin{gathered}
x08egin{array}{*{20}{c}}
{sec heta = }&{frac{{,,,13}}{x08egin{gathered}
,,,12,, hfill \
^{ imes 3} downarrow hfill \
end{gathered} }} \
{}&{,,,36}
end{array} hfill \
5{ ext{ unit}} o 5 imes 3 = 15{ ext{ m}} hfill \
end{gathered} ]
[#42] As observed from the top a light house, 120√3 m above the sea level, the angle of depression of a ship sailing towards it changes from 30° to 60°. The distance travelled by the during the period of observation is:
Correct Answer
(A) 240 m
Explanation
Solution: $$eqalign{
& sqrt 3 o 120sqrt 3 cr
& 1{ ext{u}} o 120 cr
& 2{ ext{u}} o 120 imes 2 = 240{ ext{m}} cr} $$
[#43] From the top of hill 240 m high, the angles the angles of depression of the top and bottom of a pole are 30° and 60°, respectively. The difference (in m) between the height of the pole and its distance from the hill is:
Correct Answer
(D) 80(2 - √3)
Explanation
Solution: $$eqalign{
& 3 o 240 cr
& sqrt 3 o frac{{240}}{3} imes sqrt 3 = 80sqrt 3 cr
& { ext{Height of the pole}} cr
& = 160 - 80sqrt 3 cr
& = 80left( {2 - sqrt 3 }
ight) cr} $$
[#44] A person was standing on a road near a mall. He was 1215 m away from the mall and able to see the top of the mall from the road in such a ways that the top of a tree, which is in between him and the mall, was exactly in line of sight with the top of the mall. The tree height is 20 m and it is 60 m away from him. How tall (in m) is the mall?
Correct Answer
(C) 405
Explanation
Solution: $$eqalign{
& an heta = frac{{20}}{{60}} = frac{1}{3} cr
& { ext{Now }} an heta = frac{h}{{1215}} cr
& frac{1}{3} = frac{h}{{1215}} cr
& h = 405 cr} $$
[#45] The distance between two parallel poles is 40√3 m. The angle of depression of the top of the second pole when seen from the top of first pole is 30°. What will be the height of second tower if the first pole is 100 m long
Correct Answer
(D) 60 m
Explanation
Solution: $$eqalign{
& an {30^ circ } = frac{x}{{40sqrt 3 }} cr
& frac{1}{{sqrt 3 }} = frac{x}{{40sqrt 3 }} cr
& x = 40 cr
& { ext{So, }}y = 100 - 40 = 60 cr} $$