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?
Understanding the Pillar Height and Distance Problem
This problem involves two pillars of the same height on opposite sides of a road. We are given the width of the road and the angles of elevation to the tops of the pillars from a single point on the road between them. Our goal is to find the distance of this point from the base of one of the pillars using principles of trigonometry, specifically the tangent function.
Setting up the Geometry
Let's visualize the scenario. We have two pillars, A and B, with the same height. Let the height of each pillar be \(h\) meters. The road is 40 meters wide. Let the point on the road between the pillars be P. We need to find the distance of point P from the foot of pillar A. Let this distance be \(x\) meters.
Since the total width of the road is 40 meters and the point P is between the pillars, the distance of point P from the foot of pillar B will be \((40 - x)\) meters.
We are given the angles of elevation from point P to the tops of pillars A and B.
- Angle of elevation to the top of pillar A from P is \(30^\circ\).
- Angle of elevation to the top of pillar B from P is \(45^\circ\).
Using Trigonometry to Relate Height and Distance
We can use the tangent function in trigonometry, which relates the angle of elevation, the opposite side (height of the pillar), and the adjacent side (distance from the point to the base of the pillar).
For pillar A:
In the right-angled triangle formed by pillar A, the point P, and the base of pillar A, the height of the pillar is the opposite side, and the distance \(x\) is the adjacent side to the angle of elevation \(30^\circ\).
So, we have:
\(\tan(30^\circ) = \frac{\text{Height of pillar A}}{\text{Distance of P from A}}\)
\(\tan(30^\circ) = \frac{h}{x}\)
We know that \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\). Substituting this value, we get:
\(\frac{1}{\sqrt{3}} = \frac{h}{x}\)
This gives us an expression for the height \(h\) in terms of \(x\):
\(h = \frac{x}{\sqrt{3}}\) (Equation 1)
For pillar B:
In the right-angled triangle formed by pillar B, the point P, and the base of pillar B, the height of the pillar is the opposite side, and the distance \((40 - x)\) is the adjacent side to the angle of elevation \(45^\circ\).
So, we have:
\(\tan(45^\circ) = \frac{\text{Height of pillar B}}{\text{Distance of P from B}}\)
\(\tan(45^\circ) = \frac{h}{40 - x}\)
We know that \(\tan(45^\circ) = 1\). Substituting this value, we get:
\(1 = \frac{h}{40 - x}\)
This gives us another expression for the height \(h\):
\(h = 40 - x\) (Equation 2)
Solving for the Distance
Since the pillars are of the same height, we can equate the two expressions for \(h\) from Equation 1 and Equation 2:
\(\frac{x}{\sqrt{3}} = 40 - x\)
Now, we need to solve this equation for \(x\). Multiply both sides by \(\sqrt{3}\):
\(x = \sqrt{3}(40 - x)\)
Distribute \(\sqrt{3}\) on the right side:
\(x = 40\sqrt{3} - x\sqrt{3}\)
Move the term \(x\sqrt{3}\) from the right side to the left side. Remember that when you move a term across the equals sign, you change its sign:
\(x + x\sqrt{3} = 40\sqrt{3}\)
Factor out \(x\) from the terms on the left side:
\(x(1 + \sqrt{3}) = 40\sqrt{3}\)
Now, isolate \(x\) by dividing both sides by \((1 + \sqrt{3})\):
\(x = \frac{40\sqrt{3}}{1 + \sqrt{3}}\)
To simplify this expression and remove the square root from the denominator, we can rationalize the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator, which is \((\sqrt{3} - 1)\):
\(x = \frac{40\sqrt{3}}{1 + \sqrt{3}} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1}\)
In the numerator, multiply \(40\sqrt{3}\) by \((\sqrt{3} - 1)\):
\(40\sqrt{3} \times \sqrt{3} = 40 \times 3 = 120\)
\(40\sqrt{3} \times -1 = -40\sqrt{3}\)
So, the numerator becomes \(120 - 40\sqrt{3}\).
In the denominator, we use the difference of squares formula: \((a+b)(a-b) = a^2 - b^2\). Here \(a = \sqrt{3}\) and \(b = 1\).
\((\sqrt{3} + 1)(\sqrt{3} - 1) = (\sqrt{3})^2 - (1)^2 = 3 - 1 = 2\)
So, the expression for \(x\) becomes:
\(x = \frac{120 - 40\sqrt{3}}{2}\)
Now, divide both terms in the numerator by 2:
\(x = \frac{120}{2} - \frac{40\sqrt{3}}{2}\)
\(x = 60 - 20\sqrt{3}\)
Let's check the options provided. The options are in a slightly different format. Let's factor out 20 from our result:
\(x = 20(3 - \sqrt{3})\)
This matches one of the given options. The distance of the point from the foot of pillar A is \(20(3 - \sqrt{3})\) meters.
| Parameter | Value/Expression |
|---|---|
| Height of pillars (h) | Same for A and B |
| Road Width | 40 m |
| Distance from P to A | \(x\) m |
| Distance from P to B | \((40 - x)\) m |
| Angle of elevation to A | \(30^\circ\) |
| Angle of elevation to B | \(45^\circ\) |
| \(\tan(30^\circ)\) | \(\frac{1}{\sqrt{3}}\) |
| \(\tan(45^\circ)\) | \(1\) |
| Equation from Pillar A | \(h = \frac{x}{\sqrt{3}}\) |
| Equation from Pillar B | \(h = 40 - x\) |
| Final Solution for x | \(20(3 - \sqrt{3})\) m |
Summary of the Solution Steps
- Identify the known values: road width (40m), angles of elevation (\(30^\circ\) and \(45^\circ\)).
- Assign variables: height of pillars (\(h\)), distance from point P to pillar A (\(x\)), distance from point P to pillar B (\(40-x\)).
- Use the tangent function: Set up equations relating \(h\), \(x\), and the angles using \(\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}\).
- Equate the expressions for \(h\) since the heights are equal.
- Solve the resulting equation for \(x\). This involves algebraic manipulation and rationalizing the denominator.
- Simplify the expression for \(x\) to match the format of the options.
Revision Table: Key Trigonometric Values
| Angle (\(\theta\)) | \(\sin(\theta)\) | \(\cos(\theta)\) | \(\tan(\theta)\) |
|---|---|---|---|
| \(0^\circ\) | 0 | 1 | 0 |
| \(30^\circ\) | \(\frac{1}{2}\) | \(\frac{\sqrt{3}}{2}\) | \(\frac{1}{\sqrt{3}}\) |
| \(45^\circ\) | \(\frac{1}{\sqrt{2}}\) | \(\frac{1}{\sqrt{2}}\) | 1 |
| \(60^\circ\) | \(\frac{\sqrt{3}}{2}\) | \(\frac{1}{2}\) | \(\sqrt{3}\) |
| \(90^\circ\) | 1 | 0 | Undefined |
Additional Information: Angles of Elevation and Depression
The angle of elevation is the angle formed by the line of sight to an object above the horizontal level. Imagine looking straight ahead (horizontal) and then tilting your head up to see something tall; the angle your line of sight makes with the horizontal is the angle of elevation.
The angle of depression is the angle formed by the line of sight to an object below the horizontal level. Imagine looking straight ahead and then tilting your head down to see something on the ground; the angle your line of sight makes with the horizontal is the angle of depression. Angle of elevation and angle of depression to the same point from different locations are often equal (alternate interior angles when horizontal lines are parallel).
In problems involving heights and distances, angles of elevation and depression are typically used in conjunction with trigonometric ratios (sine, cosine, and tangent) to find unknown lengths or angles in right-angled triangles.
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Important Questions from Heights and Distances
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