On the same side of a tower, two objects are located. Observed from the top of the tower, their angles of depression are 45° and 60°. If the height of the tower is 600 m, the distance between the objects is approximately equal to :
272 m
284 m
288 m
254 m
Explanation:
Let DC be the tower and A and B be the objects as shown above.
Given that DC = 600 m , $\angle$DAC = 45°, $\angle$DBC = 60°
tan60°=$\dfrac{DC}{BC}$
=>√3=$\dfrac{600}{BC}$
=>BC=$\dfrac{600}{\sqrt{3}}$ ⋯(1)
tan45°=$\dfrac{DC}{AC}$
=>1=$\dfrac{600}{AC}$
=>AC=600 ⋯(2)
Distance between the objects
= AB = (AC - BC)
=600−$\dfrac{600}{\sqrt{3}} $ [∵ from (1) and (2)]
=600 $\left( 1-\dfrac{1}{\sqrt{3}} \right) $
=600 $\left( \dfrac{\sqrt{3}-1}{\sqrt{3}} \right) $
=600 $\left( \dfrac{\sqrt{3}-1}{\sqrt{3}} \right) \times \dfrac{\sqrt{3}}{\sqrt{3}}$
=$\dfrac{600\sqrt{3}(\sqrt{3}−1)}{3} $
=200√3(√3−1)
=200(3−√3)
=200(3−1.73)
=254 m
Let DC be the tower and A and B be the objects as shown above.
Given that DC = 600 m , $\angle$DAC = 45°, $\angle$DBC = 60°
tan60°=$\dfrac{DC}{BC}$
=>√3=$\dfrac{600}{BC}$
=>BC=$\dfrac{600}{\sqrt{3}}$ ⋯(1)
tan45°=$\dfrac{DC}{AC}$
=>1=$\dfrac{600}{AC}$
=>AC=600 ⋯(2)
Distance between the objects
= AB = (AC - BC)
=600−$\dfrac{600}{\sqrt{3}} $ [∵ from (1) and (2)]
=600 $\left( 1-\dfrac{1}{\sqrt{3}} \right) $
=600 $\left( \dfrac{\sqrt{3}-1}{\sqrt{3}} \right) $
=600 $\left( \dfrac{\sqrt{3}-1}{\sqrt{3}} \right) \times \dfrac{\sqrt{3}}{\sqrt{3}}$
=$\dfrac{600\sqrt{3}(\sqrt{3}−1)}{3} $
=200√3(√3−1)
=200(3−√3)
=200(3−1.73)
=254 m