Length of the shadow of a 15 meter high pole is 5√3 meters at 7 o'clock in the morning. Then, what is the angle of elevation of the Sun rays with the ground at the time?
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From the figure above ,
Height of the pole ( AB ) = 15 m
length the of the shadow ( BC ) = 5√3 m
Angle of elevation = x°
In triangle ABC ,
<B = 90°
tan <x = AB/BC
= 15/( 5√3 )
= 3/√3
= ( √3 × √3 )/( √3 )
= √3
tan x = tan 60°
Therefore ,
x = 60°
I hope this helps you.
: ,)
Height of the pole ( AB ) = 15 m
length the of the shadow ( BC ) = 5√3 m
Angle of elevation = x°
In triangle ABC ,
<B = 90°
tan <x = AB/BC
= 15/( 5√3 )
= 3/√3
= ( √3 × √3 )/( √3 )
= √3
tan x = tan 60°
Therefore ,
x = 60°
I hope this helps you.
: ,)
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Answered by
49
Answer:
The height of the pole (AB)=15m
The length of the shadow(BC)=5√3
angle B=90°
angle of elevation=x°
Step-by-step explanation:
tanx°=AB/BC
= 15/5√3
=3/√3
=√3*√3/√3
=√3
tanx°=√3
tanx°=tan60°
therefore,x°=60°
I think so this may help you,thanks.
; )
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