ABCD is a square field with AB = x. A vertical pole OP of height 2x stands at the centre O of the square field. If ∠APO = θ, then what is cot θ equal to ?
2√2
Solving the Square Field and Vertical Pole Problem
This problem involves a combination of geometry and trigonometry. We are given a square field ABCD with side length x. A vertical pole OP of height 2x is placed at the center O of the square. We need to find the value of \(\cot \theta\), where \(\theta\) is the angle \(\angle APO\).
Understanding the Geometry of the Square
The square field is ABCD with side length AB = x. The center O of the square is the intersection point of its diagonals AC and BD. Since the pole OP is vertical, it stands perpendicular to the plane of the square field. This means that the angle between the pole (OP) and any line in the plane of the square passing through O, such as OA, OB, OC, or OD, is \(90^\circ\).
To solve the problem, we first need to find the distance from the center O to one of the vertices, say A. This distance OA is half the length of the diagonal of the square.
Let's calculate the length of the diagonal AC using the Pythagorean theorem in the right-angled triangle ABC:
\(AC^2 = AB^2 + BC^2\)
Since ABCD is a square, AB = BC = x.
\(AC^2 = x^2 + x^2 = 2x^2\)
\(AC = \sqrt{2x^2} = x\sqrt{2}\)
The distance OA is half of the diagonal AC:
\(OA = \frac{1}{2} AC = \frac{1}{2} (x\sqrt{2}) = \frac{x\sqrt{2}}{2}\)
Setting up the Trigonometry in Triangle APO
Now consider the triangle APO. P is the top of the vertical pole, O is the bottom of the pole at the center of the square, and A is a corner of the square. Since the pole OP is vertical, the angle \(\angle AOP\) is \(90^\circ\). Thus, \(\triangle APO\) is a right-angled triangle with the right angle at O.
We are given the height of the pole OP = 2x and we have calculated OA = \(\frac{x\sqrt{2}}{2}\). The angle \(\angle APO\) is given as \(\theta\).
In the right-angled triangle APO:
- The side opposite to angle \(\theta\) (\(\angle APO\)) is OA.
- The side adjacent to angle \(\theta\) (\(\angle APO\)) is OP.
- The hypotenuse is AP.
Calculating the Cotangent
We need to find \(\cot \theta\). The cotangent of an angle in a right-angled triangle is defined as the ratio of the length of the adjacent side to the length of the opposite side.
\(\cot \theta = \frac{\text{Adjacent side}}{\text{Opposite side}}\)
In \(\triangle APO\), the side adjacent to \(\theta\) is OP, and the side opposite to \(\theta\) is OA.
\(\cot \theta = \frac{OP}{OA}\)
Substitute the values we know:
\(OP = 2x\)
\(OA = \frac{x\sqrt{2}}{2}\)
\(\cot \theta = \frac{2x}{\frac{x\sqrt{2}}{2}}\)
To simplify the expression, multiply the numerator by the reciprocal of the denominator:
\(\cot \theta = 2x \times \frac{2}{x\sqrt{2}}\)
\(\cot \theta = \frac{4x}{x\sqrt{2}}\)
Cancel out the common factor x (assuming x > 0, which is true for a side length):
\(\cot \theta = \frac{4}{\sqrt{2}}\)
To rationalize the denominator, multiply the numerator and denominator by \(\sqrt{2}\):
\(\cot \theta = \frac{4}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}\)
\(\cot \theta = \frac{4\sqrt{2}}{2}\)
\(\cot \theta = 2\sqrt{2}\)
Therefore, the value of \(\cot \theta\) is \(2\sqrt{2}\).
Summary of Steps:
- Identify the geometric shape (square) and key points (vertices A, B, C, D, center O, pole top P).
- Determine the location of the center O and its relationship to the vertices (half the diagonal).
- Calculate the length of the diagonal of the square.
- Calculate the distance OA from the center to a vertex.
- Recognize the right-angled triangle APO formed by the vertex, center, and pole top.
- Identify the angle \(\theta\) and the sides opposite and adjacent to it in \(\triangle APO\).
- Use the definition of cotangent (\(\frac{\text{Adjacent}}{\text{Opposite}}\)) to set up the expression for \(\cot \theta\).
- Substitute the known lengths of OP and OA into the expression.
- Simplify the expression to find the final value of \(\cot \theta\).
| Quantity | Value | How it was obtained |
|---|---|---|
| Side length of square (x) | x | Given |
| Height of pole (OP) | 2x | Given |
| Diagonal of square (AC) | \(x\sqrt{2}\) | Pythagorean theorem in \(\triangle ABC\) |
| Distance from center to vertex (OA) | \(\frac{x\sqrt{2}}{2}\) | Half of diagonal AC |
| Angle \(\angle APO\) | \(\theta\) | Given |
| \(\cot \theta\) | \(\frac{OP}{OA}\) | Definition of cotangent in \(\triangle APO\) |
| Calculated \(\cot \theta\) | \(2\sqrt{2}\) | Substitution and simplification |
Revision Table: Key Concepts
| Concept | Description | Relevance to Problem |
|---|---|---|
| Square Properties | Four equal sides, four \(90^\circ\) angles, diagonals bisect each other at \(90^\circ\). | Used to find diagonal length and confirm O is center. |
| Pythagorean Theorem | In a right triangle, \(a^2 + b^2 = c^2\). | Used to calculate the diagonal length of the square. |
| Center of a Square | Intersection of diagonals; equidistant from all vertices. | Defines point O and allows calculating OA. |
| Vertical Line/Pole | Perpendicular to the horizontal plane it stands on. | Ensures \(\angle AOP = 90^\circ\), making \(\triangle APO\) a right triangle. |
| Trigonometric Ratios | Relationships between angles and side lengths in right triangles (SOH CAH TOA, and their reciprocals). | Used specifically \(\cot \theta = \frac{\text{Adjacent}}{\text{Opposite}}\) in \(\triangle APO\). |
Additional Information: Trigonometry and Geometry Review
Diagonal of a Square: For a square with side length 's', the diagonal 'd' can always be found using the Pythagorean theorem. Consider a right triangle formed by two sides and the diagonal: \(s^2 + s^2 = d^2\), which simplifies to \(2s^2 = d^2\). Taking the square root gives \(d = \sqrt{2s^2} = s\sqrt{2}\). The distance from the center to a vertex is always half the diagonal, i.e., \(\frac{s\sqrt{2}}{2}\).
Trigonometric Ratios (SOH CAH TOA and reciprocals):
- Sine (\(\sin \theta\)) = Opposite / Hypotenuse
- Cosine (\(\cos \theta\)) = Adjacent / Hypotenuse
- Tangent (\(\tan \theta\)) = Opposite / Adjacent
- Cosecant (\(\csc \theta\)) = 1 / \(\sin \theta\) = Hypotenuse / Opposite
- Secant (\(\sec \theta\)) = 1 / \(\cos \theta\) = Hypotenuse / Adjacent
- Cotangent (\(\cot \theta\)) = 1 / \(\tan \theta\) = Adjacent / Opposite
In this problem, since we had the lengths of the sides adjacent and opposite to the angle \(\theta\) in the right triangle APO, using the cotangent ratio was the direct way to find the required value.
Similar Questions
If the quadrilateral has an inscribed circle, then the sum of a pair of opposite sides equals:
ABCD is a cyclic quadrilateral. The bisectors of the angles A, B, C and D cut the circle at P, Q, R and S respectively. What is ∠PQR + ∠RSP equal to?
What is the magnitude (in radian) of the interior angle of a regular pentagon?
If the sum of all interior angles of a regular polygon is twice the sum of all its exterior angles then the polygon is
OABC is a rhombus whose three vertices lie on a circle with centre at O. If the area of the rhombus is 32√3 square cm, then what is the radius of the circle ?
A square is inscribed in a right angled triangle with legs p and q and has a common right angle with triangle. The diagonal of the square is given by
ABCDA is a con-cyclic quadrilateral of a circle ABCD with radius r and centre at O. If AB is the diameter and CD is parallel and half of AB and if the circle completes one rotation about the centre O, then the locus of the middle point of CD is a circle of radius:
The diagonals of a rhombus are of length 20 cm and 48 cm. What is the length of a side of the rhombus?
The area of a regular hexagon of side ‘a’ is equal to
Two parallel sides of a trapezium are 29 cm and 21 cm. Non-parallel sides are equal and each is of length 8.5 cm. What is the area of the trapezium?
Important Questions from Quadrilaterals
The ratio between the length and breadth of a rectangular park is 3 : 2. If a man cycling along the boundary at the speed of 12 km per hour completes one round in 8 minutes, then the area of the park in square meter will be
The side of a rhombus is 26 cm. The length of one of its diagonals is 20 cm. The sum of the lengths of the diagonals of this rhombus is equal to the perimeter of a rectangle. If the difference between the length and breadth of the rectangle is 6 cm, then what is the area of the rectangle?
PQRS is a cyclic quadrilateral. If ∠P is 4 times ∠R, and ∠S is 3 times ∠Q, then the average of ∠Q and ∠R is:
ABCD is a trapezium in which AB || DC and DC is perpendicular to BC. If ∠DAB = 110°, then ∠ABC - ∠ADC =_____.
If the quadrilateral has an inscribed circle, then the sum of a pair of opposite sides equals:
