Question

If the angle of elevation of a cloud from a point h metres above a lake is a. and the angle of depression of its reflection in the lake be ß, prove that the distance (x) of the cloud from the point 2h seca of observation is tanß-tana Find x if α = 30°, B = 45° and h=250​

Answer 1

Answer:

Step-by-step explanation:

Complete Question is:

the distance of the cloud from the point of observation is 2hSeca/(Tanb - Tana)

Let say Height of Cloud from Lake C

Then Vertical height of Cloud from observation point = C - h => Vertical height of Cloud image in lake

from observation point = C + h

Tana (Ch)/( Horizontal Distance) Tanb= (C + h)/( Horizontal Distance)

=> Tana/Tanb = (C-h)/(C + h)

=> CTana + hTana = CTanb - hTanb

=> C(Tanb - Tana) = h(Tana + Tanb)

=> C = h(Tana + Tanb)/(Tanb - Tana)

Sina (Ch)/ the distance of the cloud from the point of observation

=> the distance of the cloud from the

point of observation is = (C-h)/Sina

putting Ch(Tana + Tanb)/(Tanb - Tana) = (h(Tana + Tanb)/(Tanb - Tana) - h)/Sina

=h(Tana + Tanb - Tanb + tana)/Sina(Tanb

Tana)

= 2hTana/Sina(Tanb - Tana)

= 2hSina/CosaSina(Tanb - Tana)

= 2h/Cosa(Tanb - Tana)

= 2hSeca/(Tanb - Tana)

QED

Proved