NTA Abhyas 2020
- Exam
- NTA Abhyas
- Year
- 2020
- Subject
- Mathematics
- Chapter
- Heights and Distances
NTA Abhyas 2020 · Mathematics · Heights and Distances
A tower subtends an angle of 6 0 ^ o at a point on the same level...
A tower subtends an angle of at a point on the same level as the foot of the tower and at a second point just meters above the first point the angle of depression of the foot of the tower is The height of the tower is (in meters)
Options
- A\dfrac{10}{\sqrt{3}} \left(2 -\sqrt{3}\right)
- B10\sqrt{3} \left(2 -\sqrt{3}\right)
- C\dfrac{10}{\sqrt{3}} \left(2 +\sqrt{3}\right)
- D10\sqrt{3} \left(2 +\sqrt{3}\right)
Correct answer
D. 10\sqrt{3} \left(2 +\sqrt{3}\right)
Step-by-step solution
In the figure is the vertical tower and are the two points
Now, in right angled
… \left(1\right)\Delta BCE, \dfrac{BE}{EC} = tan {15}^{o}∴ \dfrac{10}{x} = tan {15}^{o} ⇒ x = \dfrac{10}{tan {15}^{o}}… \left(2\right)
From \left(1\right) we get
h = \dfrac{10}{tan {15}^{o}} \times \sqrt{3}= 10\sqrt{3} \left(2 + \sqrt{3}\right)& meters
Now, in right angled
… \left(1\right)\Delta BCE, \dfrac{BE}{EC} = tan {15}^{o}∴ \dfrac{10}{x} = tan {15}^{o} ⇒ x = \dfrac{10}{tan {15}^{o}}… \left(2\right)
From \left(1\right) we get
h = \dfrac{10}{tan {15}^{o}} \times \sqrt{3}= 10\sqrt{3} \left(2 + \sqrt{3}\right)& meters