Two persons are on either sides 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 the speed of the car?
24√3 km/hr
None of these
24/√3km/hr
20√3/3 km/hr
Explanation:
Let BD be the tower and A and C be the positions of the persons.
Given that BD = 50 m, $\angle$ BAD = 30°, $\angle$BCD = 60°
From the right $\triangle$ ABD,
tan30°=$\dfrac{BD}{BA}$
=>$\dfrac{1}{\sqrt{3}}=\dfrac{50}{BA}$
⇒BA=50√3
From the right $\triangle$ CBD,
tan60°=$\dfrac{BD}{BC}$
=>√3=$\dfrac{50}{BC}$
=>BC=$\dfrac{50}{\sqrt{3}}=\dfrac{50 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}=\dfrac{50\sqrt{3}}{3}$
Distance between the two persons
= AC = BA + BC
=50√3+$\dfrac{50\sqrt{3}}{3}$
=√3 $\left( 50+\dfrac{50}{3} \right) $
=$\dfrac{200\sqrt{3}}{3}$ m
i.e., the distance travelled by the car in 10 seconds =$\dfrac{200\sqrt{3}}{3}$ m
Speed of the car =$\dfrac{Distance}{Time}$
=$\dfrac{\left( \dfrac{200\sqrt{3}}{3}\right)}{10}= \dfrac{20\sqrt{3}}{3}$ meter/second
= $\dfrac{20\sqrt{3}}{3} \times \dfrac{18}{5}$km/hr==24√3 km/hr
Let BD be the tower and A and C be the positions of the persons.
Given that BD = 50 m, $\angle$ BAD = 30°, $\angle$BCD = 60°
From the right $\triangle$ ABD,
tan30°=$\dfrac{BD}{BA}$
=>$\dfrac{1}{\sqrt{3}}=\dfrac{50}{BA}$
⇒BA=50√3
From the right $\triangle$ CBD,
tan60°=$\dfrac{BD}{BC}$
=>√3=$\dfrac{50}{BC}$
=>BC=$\dfrac{50}{\sqrt{3}}=\dfrac{50 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}=\dfrac{50\sqrt{3}}{3}$
Distance between the two persons
= AC = BA + BC
=50√3+$\dfrac{50\sqrt{3}}{3}$
=√3 $\left( 50+\dfrac{50}{3} \right) $
=$\dfrac{200\sqrt{3}}{3}$ m
i.e., the distance travelled by the car in 10 seconds =$\dfrac{200\sqrt{3}}{3}$ m
Speed of the car =$\dfrac{Distance}{Time}$
=$\dfrac{\left( \dfrac{200\sqrt{3}}{3}\right)}{10}= \dfrac{20\sqrt{3}}{3}$ meter/second
= $\dfrac{20\sqrt{3}}{3} \times \dfrac{18}{5}$km/hr==24√3 km/hr