1
1
the angle of depression of the top and the foot of a 9 m tall building from the top of a multi-storeyed building are $30^\circ$ and $60^\circ$ respectively. Find the height of the multi-storeyed building and the distance between the two buildings.
the angle of depression of the top and the foot of a 9 m tall building from the top of a multi-storeyed building are $30^\circ$ and $60^\circ$ respectively. Find the height of the multi-storeyed building and the distance between the two buildings. (Use $\sqrt{3} = 1.73$)
This Question has 1 answers.
Let the height of the multi-storeyed building be H meters.
Let the distance between the two buildings be } d meters.
$ \text{Given:} $
$ \text{Height of the smaller building } = 9 \text{ m} $
Angle of depression to the top of the smaller building = \theta_1
Angle of depression to the foot of the smaller building = \theta_2
$ \text{Using } \sqrt{3} \approx 1.73 $
From the top of the multi-storeyed building, two right-angled triangles are formed.
$ \text{In } \triangle ABC, \tan \theta_1 = \frac{(H - 9)}{d} $
$ \Rightarrow H - 9 = d \tan \theta_1 $
$ \Rightarrow H = 9 + d \tan \theta_1 $
$ \text{In } \triangle ABD, \tan \theta_2 = \frac{H}{d} $
$ \Rightarrow H = d \tan \theta_2 $
$ \text{Equating both equations:} $
$ d \tan \theta_2 = 9 + d \tan \theta_1 $
$ d (\tan \theta_2 - \tan \theta_1) = 9 $
$ d = \frac{9}{\tan \theta_2 - \tan \theta_1} $
$ \text{Now, substituting } d \text{ in } H = d \tan \theta_2, $
$ H = \frac{9 \tan \theta_2}{\tan \theta_2 - \tan \theta_1} $
Thus, the height of the multi-storeyed building is H meters, and the distance between the two buildings is } d meters.
Let the distance between the two buildings be } d meters.
$ \text{Given:} $
$ \text{Height of the smaller building } = 9 \text{ m} $
Angle of depression to the top of the smaller building = \theta_1
Angle of depression to the foot of the smaller building = \theta_2
$ \text{Using } \sqrt{3} \approx 1.73 $
From the top of the multi-storeyed building, two right-angled triangles are formed.
$ \text{In } \triangle ABC, \tan \theta_1 = \frac{(H - 9)}{d} $
$ \Rightarrow H - 9 = d \tan \theta_1 $
$ \Rightarrow H = 9 + d \tan \theta_1 $
$ \text{In } \triangle ABD, \tan \theta_2 = \frac{H}{d} $
$ \Rightarrow H = d \tan \theta_2 $
$ \text{Equating both equations:} $
$ d \tan \theta_2 = 9 + d \tan \theta_1 $
$ d (\tan \theta_2 - \tan \theta_1) = 9 $
$ d = \frac{9}{\tan \theta_2 - \tan \theta_1} $
$ \text{Now, substituting } d \text{ in } H = d \tan \theta_2, $
$ H = \frac{9 \tan \theta_2}{\tan \theta_2 - \tan \theta_1} $
Thus, the height of the multi-storeyed building is H meters, and the distance between the two buildings is } d meters.
Similar Questions
- There are three taps A, B and C in a tank. They can fill the tank in 25 hrs, 20 hrs and 10 hrs respectively. At first all of them are opened simultaneously. Then after 1 hrs, tap C is closed and tap A and B are kept running. After the 4th hour, tap B is a
- In a hotel, the rooms are numbered from 101 to 130 on the first floor, 221 to 260 on the second floor and 306 to 345 on the third floor. In the month of June this year, the room occupancy was 60% on the first floor, 40% on the second floor and 75% on the
- Nidhi received simple interest off Rs. 1200 when invested Rs x at 6% p.a. and Rs y at 5% p.a. for 1 year. Had she invested Rs. x at 3% p.a. and Rs. y at 8% for that year, she would have received simple interest of Rs 1,260. Find the values Rs. x an y.
- A and B can complete a piece of work in 80 and 120 days respectively. They together start the work but A left after 20 days. After another 12 days C joined B and now they complete the work in 28 more days. In how many days C can complete the work, working
- S, T and U can complete a work in 40, 48 and 60 days respectively. They received Rs 10800 to complete the work. They begin the work together but T left 2 days before the completion of the work and U left 5 days before the completion of the work. S has com