The angle of elevation of the top of a tall building from the points M and N at the distances of 72 m and 128 m, respectilvely, from the base of the building and in the same straight line with it, are complementary. The height of the building (in m) is:
96
Solving for Building Height with Complementary Angles of Elevation
This problem involves trigonometry, specifically the concept of angles of elevation and complementary angles, applied to a right-angled triangle formed by the building, the ground, and the line of sight to the top of the building.
Let's define the terms:
- Angle of Elevation: The angle between the horizontal line from the observer's eye to an object and the line of sight to the object, when the object is above the horizontal line.
- Complementary Angles: Two angles are complementary if their sum is 90 degrees ($90^\circ$).
Setting up the Problem Geometry
Consider a tall building. Let the height of the building be \(H\) meters. Let the base of the building be point B and the top of the building be point T. Points M and N are on the ground, in the same straight line with the base B.
The distance from the base B to point M is given as 72 m. The distance from the base B to point N is given as 128 m.
Let the angle of elevation of the top of the building (T) from point M be \(\theta_M\) and from point N be \(\theta_N\).
The problem states that the angles of elevation from M and N are complementary. Therefore, \(\theta_M + \theta_N = 90^\circ\).
Using Trigonometry to Relate Angles and Height
We have two right-angled triangles: \(\triangle TBM\) and \(\triangle TBN\), both right-angled at B.
In \(\triangle TBM\), the opposite side to angle \(\theta_M\) is the height \(H\), and the adjacent side is the distance BM (72 m). Using the tangent ratio:
\(\tan(\theta_M) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{H}{72}\)
In \(\triangle TBN\), the opposite side to angle \(\theta_N\) is the height \(H\), and the adjacent side is the distance BN (128 m). Using the tangent ratio:
\(\tan(\theta_N) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{H}{128}\)
Applying the Complementary Angle Condition
We know that \(\theta_M + \theta_N = 90^\circ\). This implies \(\theta_M = 90^\circ - \theta_N\).
Substitute this into the equation for \(\tan(\theta_M)\):
\(\tan(90^\circ - \theta_N) = \frac{H}{72}\)
Using the trigonometric identity \(\tan(90^\circ - x) = \cot(x)\), we get:
\(\cot(\theta_N) = \frac{H}{72}\)
We also know that \(\cot(\theta_N) = \frac{1}{\tan(\theta_N)}\). From the equation for \(\tan(\theta_N)\), we have \(\tan(\theta_N) = \frac{H}{128}\). So, \(\cot(\theta_N) = \frac{128}{H}\).
Solving for the Height of the Building
Now we can equate the two expressions for \(\cot(\theta_N)\):
\(\frac{H}{72} = \frac{128}{H}\)
Multiply both sides by \(H \times 72\) to clear the denominators:
\(H \times H = 72 \times 128\)
\(H^2 = 72 \times 128\)
Now, let's calculate the product:
\(H^2 = 9216\)
To find \(H\), take the square root of both sides:
\(H = \sqrt{9216}\)
To find the square root, we can factorize the numbers:
\(72 = 8 \times 9 = 2^3 \times 3^2\)
\(128 = 2^7\)
\(H^2 = (2^3 \times 3^2) \times 2^7 = 2^{3+7} \times 3^2 = 2^{10} \times 3^2\)
\(H = \sqrt{2^{10} \times 3^2} = \sqrt{(2^5)^2 \times 3^2} = 2^5 \times 3\)
\(H = 32 \times 3\)
\(H = 96\)
So, the height of the building is 96 meters.
Verification of the Result
Let \(H = 96\). \(\tan(\theta_M) = \frac{96}{72} = \frac{4}{3}\) \(\tan(\theta_N) = \frac{96}{128} = \frac{3}{4}\)
We see that \(\tan(\theta_N) = \frac{1}{\tan(\theta_M)}\). This means \(\tan(\theta_N) = \cot(\theta_M)\). Since \(\cot(\theta_M) = \tan(90^\circ - \theta_M)\), we have \(\tan(\theta_N) = \tan(90^\circ - \theta_M)\). This implies \(\theta_N = 90^\circ - \theta_M\), or \(\theta_M + \theta_N = 90^\circ\). The angles are indeed complementary.
| Concept | Formula/Relation |
|---|---|
| Tangent of angle in right triangle | \(\tan(\theta) = \frac{\text{Opposite Side}}{\text{Adjacent Side}}\) |
| Complementary Angles | If \(\alpha + \beta = 90^\circ\), then \(\alpha\) and \(\beta\) are complementary. |
| Trigonometric Identity for Complementary Angles | \(\tan(90^\circ - \theta) = \cot(\theta)\) |
| Relation between tan and cot | \(\cot(\theta) = \frac{1}{\tan(\theta)}\) |
Revision Table: Angle of Elevation Problem
| Step | Description | Calculation/Formula |
|---|---|---|
| 1 | Define variables (Height H, distances 72m, 128m) | |
| 2 | Write tangent equations for each point | \(\tan(\theta_M) = H/72\), \(\tan(\theta_N) = H/128\) |
| 3 | Use complementary angle relation | \(\theta_M + \theta_N = 90^\circ \implies \theta_M = 90^\circ - \theta_N\) |
| 4 | Apply identity \(\tan(90^\circ - \theta) = \cot(\theta)\) | \(\tan(\theta_M) = \cot(\theta_N)\) |
| 5 | Substitute tangent expressions | \(\frac{H}{72} = \frac{128}{H}\) |
| 6 | Solve for H | \(H^2 = 72 \times 128 \implies H = \sqrt{9216} = 96\) |
Additional Information: Applications of Angle of Elevation
Angles of elevation are widely used in various fields:
- Surveying: To determine the height of buildings, mountains, towers, etc.
- Navigation: Used by pilots and sailors to determine their position relative to ground or sea level.
- Astronomy: To measure the altitude of celestial bodies above the horizon.
- Engineering: In civil engineering for construction and structural design.
- Photography: To compose shots involving vertical elements like buildings or trees.
Understanding angle of elevation and depression is fundamental in solving problems related to heights and distances using trigonometry.
Similar Questions
A person standing at a distance looks at a building having a height of 1000 metres. The angle between the top of the building and the ground is 30°. At what approximate distance (in metres) is the person standing away from the building.
Two ships are on the opposite of a light house such that all three of them are collinear. The angles of depression of the two ships from the top of the light house are 30° and 60°. If the ships are 230√3 m apart, then find the height of the light house (in m).
- A pole 23 m long reaches a window which is 3 \(\sqrt5\) m above the ground on one side of a street. Keeping its foot at the same point, the pole is turned to the other side of the street to reach a window 4 \(\sqrt15\) m high. What is the width (in m) of the street?
A person from the top of a hill observes a vehicle moving towards him at a uniform speed. It takes 10 minutes for the angle of depression to change from 45° to 60°. After this the time required by the vehicle to reach the bottom of the hill is
Two pillars A and B of the same height are on opposite sides of a road which is 40 m wide. The angles of elevation of the tops of the pillars A and B are 30° and 45°, respectively, at a point on the road between the pillars. What is the distance (in m) of the point from the foot of pillar A?
From the top of a hill 240 m high, the angles of depression of the top and bottom of a pole are 30° and 60°, respectively. The difference (in m) between the height of the pole and its distance from the hill is:
The length of the shadow of a vertical tower on level ground increases by 10 m when the altitude of the sun changes from 45° to 30°. The height of the tower is:
A vertical tower stands on a horizontal plane and a surmounted by a vertical flagstaff of height h. At a point on the plane, the angle of elevation of the bottom of the flagstaff is α and that of the top of the flagstaff is β. Then the height of the tower is
Two ladders AB and CD are inclined on two floors of a building as shown below, such that BC = √2 m. If the height of a floor is 4√2 m, how much minimum distance is to be covered to walk from point A to point D?
A vertical pole and a vertical tower are on the same level ground in such a way that, from the top of the pole, the angle of elevation of the top of the tower is 60° and the angle of depression of the bottom of the tower is 30°. If the height of the pole is 24 m, then find the height of the tower (in m).
Important Questions from Heights and Distances
A peacock sitting at the top of a 3 meter high pole saw a snake approaching towards pole at a distance three times of the height of the pole. Then it jumping from pole will catch the snake at what distance from the pole if both are running with same speed ?
The foot of a ladder 25 m long is 7 m from the base of the building. If the top of the ladder slips by 4 m, then by how much distance will the foot of the ladder slide?
Two hotels stand 25 m apart. One of them is 70 m high and the angle of depression of the top of other as observed from the top of this hotel is 45°. Height of the other hotel is:
If the angles of elevation of a balloon from two consecutive kilometer-stones along a straight road are 30° and 60° respectively, then the height of the balloon above the ground will be:
The angle of elevation of the top of a tower from a point 20 m away from its base is 45 °. What is the height of the tower?
