A pole stands in a park such that its shadow increases by 1 m when the angle of elevation of sun changes from 45° to 30°. Find the height of the pole.
Answers
Answer:
The height of the pole is approximately 2.73m
Step-by-step explanation:
Find the attachment below for reference.
Let the pole(AB) be of height h
When the angle of elevation is 45°, let the shadow be formed at C
Let the length of this shadow(AC) be x m
When the angle of elevation is 30°, let the shadow be formed at D
As the shadow increases by 2m
Let the length of this shadow(AD) will be x+2 m
We know that tan \thetaθ = \frac{Perpendicular}{Base}
Base
Perpendicular
Therefore in Δ ABC,
tan 45° = \frac{h}{x}
x
h
=> 1 = \frac{h}{x}
x
h
(as tan 45° = 1)
=> x = h --(i)
in Δ ABD,
tan 30° = \frac{h}{x+2}
x+2
h
=> \frac{1}{\sqrt{3} }
3
1
= \frac{h}{x+2}
x+2
h
(as tan 45° = 1)
=> x+2 = h √3
=> h + 2 = h √3 (from equation(i))
=> 2 = h √3 - h
=> 2 = h( √3 - 1)
=> h = \frac{2}{\sqrt{3} -1 }
3
−1
2
We will now rationalize the denominator.
=> h = \frac{2}{\sqrt{3} -1 }
3
−1
2
* \frac{\sqrt{3} +1}{\sqrt{3} +1 }
3
+1
3
+1
=> h = \frac{2(\sqrt{3} +1)}{(\sqrt{3} +1 )(\sqrt{3} +1) }
(
3
+1)(
3
+1)
2(
3
+1)
=> h = \frac{2(\sqrt{3} +1)}{(\sqrt{3})^{2} -1 ^{2} }
(
3
)
2
−1
2
2(
3
+1)
=> h = \frac{2(\sqrt{3} +1)}{3-1}
3−1
2(
3
+1)
=> h = \frac{2(\sqrt{3} +1)}{2 }
2
2(
3
+1)
=> h = √3 + 1
=> h = 1.732 + 1
=> h = 2.732m
Therefore, the height of the pole is approximately 2.73m