Two persons are on either side of a tower of height 50 m. The persons observers the top of the tower at an angle of elevation of 30° and 60°. If a car crosses these two persons in 10 seconds, what is
Let BD be the tower and A and C be the positions of the persons.
Given that BD = 50 m, BAD = 30°, BCD = 60°
From the right ABD,
tan 30° = BD/BA
⇒1/√3=50/BA
⇒BA=50√3
From the right CBD,
tan 60° = BD/BC
⇒√3=50/BC
⇒BC=50/√3
=(50×√3)/(√3×√3)
=(50√3)/3
Distance between the two persons
= AC = BA + BC
=50√3+((50√3)/3)
=√3(50+(50/3))
=(200√3)/3 m
i.e., the distance travelled by car in 10 seconds =(200√3)/3 m
Speed of the car =Distance/Time
=((200√3/3)/10)
=(20√3)/3 meter/second
=(20√3)/3×(18/5) km/hr
=24√3 km/hr
Workspace
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BAD = 30°, 
ABD,
