An aeroplane when 900 m high passes vertically above another aeroplane at an instant when their angles of elevation at same observing point are 60° and 45° respectively. Approximately, how many meters higher is the one than the other?
381 m
169 m
254 m
211 m
Explanation:
Let C and D be the position of the aeroplanes.
Given that CB = 900 m, $\angle$CAB = 60°, $\angle$DAB = 45°
From the right $\triangle$ ABC,
tan60°=$\dfrac{CB}{AB}$
√3=$\dfrac{900}{AB} $
AB=$\dfrac{900}{\sqrt{3}} $
=$\dfrac{900 \times \sqrt{3} }{\sqrt{3} \times \sqrt{3}} =\dfrac{900 \sqrt{3} }{3} $=300√3
rom the right $\triangle$ ABD,
tan45°=$\dfrac{DB}{AB}$
1=$\dfrac{DB}{AB}$
DB=AB=300√3
Required height
= CD = (CB-DB)
=(900−300√3)=(900−300×1.73)
=(900−519)=381 m
Let C and D be the position of the aeroplanes.
Given that CB = 900 m, $\angle$CAB = 60°, $\angle$DAB = 45°
From the right $\triangle$ ABC,
tan60°=$\dfrac{CB}{AB}$
√3=$\dfrac{900}{AB} $
AB=$\dfrac{900}{\sqrt{3}} $
=$\dfrac{900 \times \sqrt{3} }{\sqrt{3} \times \sqrt{3}} =\dfrac{900 \sqrt{3} }{3} $=300√3
rom the right $\triangle$ ABD,
tan45°=$\dfrac{DB}{AB}$
1=$\dfrac{DB}{AB}$
DB=AB=300√3
Required height
= CD = (CB-DB)
=(900−300√3)=(900−300×1.73)
=(900−519)=381 m