Question

A 30m high pole was standing at a point of the length side of a rectangle garden. If the angles of elevation of that pole form end points of that length are found 60 degree and 30 degree,find the length of that garden.​

Answer 1

The length of the rectangle garden is 20
`√(3)` i.e 34.64 meters .

Explanation:

Given as :

The height of the pole = H = 30 m

The two angles of elevation as 60° and 30°

Let The length of the rectangle garden = L m

Now, From figure

The measure from pole ground to first elevation 60° = x m

The measure from pole ground to second elevation 30° = (L + x ) m

Now, from triangle BOC

Tan Ф =
`(\textrm perpendicular)/(\textrm base)`

I.e Tan 60° =
`(\textrm 30)/(\textrm x)`

Or,
`√(3)` =
`(\textrm 30)/(\textrm x)`

or, x =
`(30)/(√(3) )`

I.e x = 10
`√(3)` meter ...1

Again From triangle BAC

Tan Ф =
`(\textrm perpendicular)/(\textrm base)`

I.e Tan 30° =
`(\textrm 30)/(\textrm L + x)`

Or,
`(1)/(√(3) )` =
`(\textrm 30)/(\textrm L + x)`

Or, L + x = 30 ×
`√(3)` ....2

Put the value of x from 1 into 2

i.e L + 10
`√(3)` meter = 30 ×
`√(3)`

or, L = 30
`√(3)` - 10
`√(3)`

Or, L = 20
`√(3)` = 34.64 meters

I.e length of garden is 20
`√(3)`

Hence The length of the rectangle garden is 20
`√(3)` i.e 34.64 meters . answer

Answer 2

Length of the garden = 69.28 meters.

Explanation:

See the attached diagram.

Let CD is the pole with a height of 30 m and the elevation of the top of the pole from the endpoints of the length A and B are 60° and 30° respectively.

So, ∠ DAC = 60° and ∠ DBC = 30°

Now, Δ ACD is a right triangle and
`\tan 60 = (CD)/(AC)`

⇒
`AC = (CD)/(\tan 60) = (30)/(\tan 60) = 17.32` meters.

Now, from Δ BCD which is a right triangle,
`\tan 30 = (CD)/(CB)`

⇒
`CB = (30)/(\tan 30) = 51.96` meters.

Hence, AB = AC + CB = 17.32 + 51.96 = 69.28 meters. (Answer)