A women is standing 30 m away from her house. An angle of elevation fr...
Given,
⇒ tan30° = AB/BC
⇒ AB = 10√3 m
Given,
⇒ tan60° = AE/ED
⇒ √3 = (AB + BE)/30
⇒ AB + BE = 30√3
⇒ BE = 30√3 – 10√3
⇒ BE = 20√3 m
⇒ AE = 20√3 + 10√3 = 30√3 m
Required height = 30√3 + 20√3 = 50√3 m
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A women is standing 30 m away from her house. An angle of elevation fr...
To find the total height of the house and the woman using the angles of elevation, we can break the problem down into two parts: calculating the height of the woman and the height of the house.
Step 1: Height of the Woman
1. Given Information:
- Distance from the woman to the house = 30 m
- Angle of elevation from the top of the woman = 30°
2. Using Trigonometry:
- Let \( h_w \) be the height of the woman.
- From the angle of elevation formula:
\[
\tan(30^\circ) = \frac{h_h - h_w}{30}
\]
- We know \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \).
- Therefore,
\[
\frac{1}{\sqrt{3}} = \frac{h_h - h_w}{30}
\]
- Rearranging gives:
\[
h_h - h_w = 30 \cdot \frac{1}{\sqrt{3}} = \frac{30}{\sqrt{3}} = 10\sqrt{3}
\]
Step 2: Height of the House
1. Given Information:
- Angle of elevation from the foot of the woman = 60°
2. Using Trigonometry:
- From the foot of the woman:
\[
\tan(60^\circ) = \frac{h_h}{30}
\]
- We know \( \tan(60^\circ) = \sqrt{3} \).
- Thus,
\[
\sqrt{3} = \frac{h_h}{30}
\]
- Rearranging gives:
\[
h_h = 30\sqrt{3}
\]
Step 3: Total Height
1. Combining Heights:
- We have two equations:
\[
h_h = 30\sqrt{3}
\]
\[
h_h - h_w = 10\sqrt{3}
\]
- Solving for \( h_w \):
\[
h_w = h_h - 10\sqrt{3} = 30\sqrt{3} - 10\sqrt{3} = 20\sqrt{3}
\]
2. Total Height:
- Total height = \( h_h + h_w = 30\sqrt{3} + 20\sqrt{3} = 50\sqrt{3} \)
Thus, the total height of the house and the woman is \( 50\sqrt{3} \) m, which corresponds to option (b).