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Question

Q1. The string of a kite is 200 metres long it makes an angle of 60° with the horizontal . Find the height of the kite (there is no slack in the string).

Solution.



let OB is string and vertical height is h that is calculated, sin60° =

AB/OB

=

h/100

    ⇒    h =

√3/2

(100)

h = 50√(3)

Q2. A person, standing on the bank of a river observes that the angle subtended by a tree on the opposite bank is 60°, when he retreates 30m from the bank, he finds the angle to be 30° , find the height of the tree and the breadth of the river .

Apply trigonometric ratio in both triangles

tan60⁰ =

BC/AB

=

h/x

  ⇒    tan30⁰ =

BC/OB

=

h/30+x

h = x√3 ----1 and h =

x+20/√3

------2

equating both equations

x√3 =

x+30/√3

  ⇒   3x = x + 30   ⇒   x = 15

h = 15√3

Q3. If the angles of elevation of the top of a tower from two points distant 10m and 20m from the base (both points are on same straight line). with it are complementary, then the height of tower is

Solution.



h = DC = √(ab)

So, height is = √(10x20)

= √200   = 10√2

Q4. An observer 1.5m tail is 28.5m away from a tower . The angle of elevation of the top of the tower from her eyes is 45^o. what is the height of the tower ?

Solution.

Let AB be the tower of height h and CD be the observer of height 1.5m at a distance of 28.5m from the tower AB . In triangle AED , we have

tan 45^o =

AE/DE

⇒    1 =

AE/28.5

⇒    AE = 28.5m

∴    h = AE + BE = AE + DC

   = (28.5 + 1.5)m = 30m

Hence, the height of the tower is 30m .

Q5. The shadow of a vertical tower on level ground increases by 10 metres, when the altitude of the sun changes from angle of elevation 45^o to 30^o . find the height of the tower , correct to one place of decimal.(take √3 = 1.73)

Solution.

In triangle CAB, we have

tan 45^o =

AB/AC

⇒    1 =

h/x

⇒    x = h ---------(i)

In triangle DAB, we have

tan 30^o =

AB/AD

⇒   

1/√3

=

h/x + 10

⇒    x+10 = √3h ----------(ii)

Substituting the value of x obtained from equation (i) and (ii) , we get

h + 10 = √3h

⇒    h(√3 - 1) = 10

⇒   

10/√3 - 1

  =

10/√3 - 1

x

√3 + 1/√3 + 1

   = 10[

√3 + 1/2

]   = 5(√3 + 1)

⇒    h = 5(1.73 + 1) = 13.65 metres

Hence, the height of the tower is 13.65 metres