Question

from a balloon vertically above a straight road the angles of depression of two cars at an instant are 30 and 45.If the car 80m apart find the height of balloon

Answer 1

answer= 40(√3-1)m

explaination; let distance b/w second car and the point from where be × ;

from the first car angle of depression is 30° from the point balloon is in the air.

height be H

so,

tan 30°=H/80-x

= 1/√3=H/ 80-x

= H = 80-x/√3. ......1

now from 2nd car;

tan 45°=H/x

= H=x. .........2

from 1 and 2,

80-x/√3=x

80-x=√3x

80=√3x+x

x(√3+1)=80

×=80/√3+1

on rationalising;

×=80(√3-1)/3-1=>x=40(√3-1)m

since,H=x by..2;

height will be =40(√3-1)m

Answer 2

Answer:

40($\sqrt{3}$+1)

Step-by-step explanation:

let AB be the height of balloon

Tan 45 =AB/CB

1=AB/CB

1. CB=AB

Tan 30=AB/BD

1/$\sqrt{3}$=AB/(BC+80)

1/$\sqrt{3}$=AB/(AB+80)                     (from 1.)

AB+80=AB$\sqrt{3}$

80=AB$\sqrt{3}$-AB

80=AB($\sqrt{3}$-1)

AB=80/$\sqrt{3}$-1

AB=80($\sqrt{3}$+1)/3-1                  (multiplied numerator and denominator by $\sqrt{3}$+1)

AB=80 ($\sqrt{3}$ +1)/2

AB=40 ($\sqrt{3}$+1)