Aptitude Height And DistancePage 2

9.	Two men on the same side of a tall building notice the angle of elevation to the top of the building to be 30o and 60o respectively. If the height of the building is known to be h =120 m, find the distance (in meters) between the two men. (a)138.56 (b)135 (c)120 (d)130.56

Answer!

Answer is: A
In the figure, A and B represent the two men and CD the tall building.
tan A = tan 30^o = DC/AC = h/AC; and
tan B = tan 60^o = DC/BC = h/BC
Now the distance between the men is
AB = x = AC − BC = (h/tan 30^o) − (h/tan 60^o)
AB = (120√3 ) − (120/√3 ) = 138.56 m.

10.	A pole of height h = 50 ft has a shadow of length l = 50.00 ft at a particular instant of time. Find the angle of elevation (in degrees) of the sun at this point of time. (a)90 (b)45 (c)60 (d)30

Answer!

Answer is: B
In the figure, BC represents the pole and AB its shadow.
tan A = BC / AB = h / l = 50 / 50.00 = 1.000
From trigonometric tables, we note that
tan A = 1.000 for A =45^o
Hence the angle of elevation of the sun at this point of time is 45^o

11.	The angle of elevation of the top of a tower 30 m high, from two points on the level ground on its opposite sides are 45o and 60o. What is the distance between the two points? (a)30 (b)51.96 (c)47.32 (d)81.96

Answer!

Answer is: C
Let OT be the tower.
Therefore, Height of tower = OT = 30 m
Let A and B be the two points on the level ground on the opposite side of tower OT Then, angle of elevation from A = TAO = 45^o and angle of elevation from B = TBO = 60^o
Distance between AB = AO + OB = x + y (say)
Now, in right triangle ATO,
tan 45^o = OT/AT = 30/x
x = 30/tan45^o = 30 m
And in right triangle BTO,
tan 60^o = OT/OB = 30/y
y = 30/tan60^o = 30/√3 = 30√3/3 = 17.32 m
Hence, the required distance = x + y = 30 + 17.32 = 47.32 m

12.	A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30°. (a)8 m (b)10 m (c)11 m (d)12 m

Answer!

Answer is: B
In Triangle ABC,
Sin 30 = AB/AC
1/2 = AB/20
AB = 10 m

13.	A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree. (a)6√3 (b)8√3 (c)8 (d)6

Answer!

Answer is: Blet us use the previous figure

Here, BC = 8 m
or, tan 30 = AB/BC
Or, 1/√3 = AB/8
Or, AB = 8/√3 = 4.61 m
Now, cos 30 = √3/2 = BC/AC = 8/AC
Or, AC = 16/√3
Height of tree = AB + BC = 8/√3 + 16/√3 = 24/√3 = 8√3

14.	A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building. (a)9.5√3 (b)9√3 (c)19√3 (d)17√3

Answer!

Answer is: C
Given that, AB = 30 m(Height of building)
DC = EG = 1.5 m(Height of building)
Angle D = 30^o
Angle E = 60^o
AF = 30 - 1.5 = 28.5 m
In Triangle AFD, tan30^o = AF/FD = 28.5/FD
Or, 1/√3 = 28.5/FD
Or, FD = 28.5/√3
In Triangle AFE, tan60^o = AF/FE = 28.5/FE
Or, √3 = 28.5/FE
Or, FE = 28.5/√3
Required distance = ED = 28.5√3 - 28.5/√3
ED = 28.5√3 - 9.5√3 = 19√3

15.	From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower. (a)20.64 m (b)14.64 m (c)20 m (d)16 m

Answer!

Answer is: B
Height of building = DB = 20 m
Angle DCB = 45^o
Angle ACB = 60^o
In Triangle DBC, tan45^o = 1 = DB/BC
Or, Db = BC = 20 m
In Triangle ABC, tan60^o = √3 = AB/BC
Or, AB = 20√3
Now, AD = AB - DB = 20√3 - 20
AD = 20(√3 - 1) = 20(1.732 - 1)
AD = 20 x 0.732 = 14.64 m

16.	The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building. (a)50/3 m (b)50 m (c)50/√3 m (d)16 m

Answer!

Answer is: A
Height of tower = AB = 50 m
Angle ACB = 60^o
Angle DBC = 30^o
In Triangle ABC, tan60^o = √3 = 50/BC
Or, BC = 50/√3

In Triangle, DCB, tan30^o = 1/√3 = DC/BC
Or, DC = 50/3

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