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).
The angle of elevation of the top of a building at a distance of 70 m from its foot on a horizontal plane is found to be 60°. Find the height of the building.
From the top of an upright pole 17.75 m high, the angle of elevation of the top of an upright tower was 60⁰. If the tower was 57.75 m tall, how far away (in m) from the foot of the pole was the foot of the tower?
From the top of an upright pole 24√3 feet high, the angle of elevation of the top of an upright tower was 60°. If the foot of the pole was 60 feet away from the foot of the tower, what tall (in feet) was the tower?
The angle of elevation of the top of a tower from the top of a building whose height is 680 m is 45° and the angle of elevation of the top of same tower from the foot of the same building is 60°. What is the height (in m) of the tower?
- The angle of elevation of the top of an unfinished tower at a point distant 78 m from its base is 30°. How much higher must the tower be raised (in m) so that the angle of elevation of the top of the finished tower at the same point will be 60º?
A kite is attached to a string. Find the length of the string (in m) when the height of the kite is 90 m and the string makes an angle of 30° with the ground.
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).
A kite flying at a height of 120 m is attached to a string which makes an angle of 60° with the horizontal. What is the length (in m) of the string?
Important Questions from Heights and Distances
Two ships are sailing in the sea on the two sides of a lighthouse. The angles of elevation of the top of the lighthouse as observed from the ships are 45 ° and 60° respectively. If the lighthouse is 81 m high, then the distance between two ships is:
The horizontal distance between two towers is 40√3 m. The angle of depression of the top of the first tower when seen from the top of the second tower is 30°. If the height of the second tower is 130 m, find the height of the first tower.
The angle of elevation of a ladder leaning against a house is 60° and the foot of the ladder is 6.5 metres from the house. The length of the ladder is
A kite is flying at a height of 50 m. If the length of the string is 100 m then the inclination of the string to the horizontal ground in degree measures is:
A. 90
B. 45
C. 60
D. 30
Two poles of the height 15 m and 20 m stand vertically upright on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.
A. 11 m
B. 12 m
C. 13 m
D. 14 m
