Problem 10 - A Complete Resource Book in Mathematics for JEE Main

Question

The angle of elevation of the top of a tree at point \(B\) due south of it is \(60^{\circ}\) and at a point \(C\) due north of it is \(30^{\circ} . D\) is a point due north of \(C\) where the angle of elevation is \(15^{\circ}\). If \(\sqrt{3}=1 \frac{8}{11}\) and \(B C \times C D=2^{3} \times 3^{2}\) \(\times 19 \times 11\), the height of the tree is (A) 33 (B) 38 (C) 57 (D) 88

Step-by-Step Solution

Answer

The height of the tree is 33.

Step 1: Understand the Problem

We have a tree and three points B, C, and D. Point B is due south, and C and D are due north. The angles of elevation of the tree from these points are 60°, 30°, and 15° respectively. We are given the product of the distances BC and CD. We need to find the height of the tree.

Step 2: Express relationships using trigonometry

Use tangent for angles of elevation, which is defined as \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \). Let the height of the tree be \( h \).From point B:\[ \tan 60^{\circ} = \sqrt{3} = \frac{h}{x} \implies h = x\sqrt{3} \]From point C:\[ \tan 30^{\circ} = \frac{1}{\sqrt{3}} = \frac{h}{y} \implies h = \frac{y}{\sqrt{3}} \]From point D:\[ \tan 15^{\circ} = 2 - \sqrt{3} = \frac{h}{z} \implies h = z(2 - \sqrt{3}) \]

Step 3: Use the relationships

We know that:- \( h = x\sqrt{3} \)- \( h = \frac{y}{\sqrt{3}} \)Let's express \( x \) and \( y \):- \( y = h\sqrt{3} \)- \( x \cdot y = 1134 \) (from \( BC \times CD \))

Step 4: Solve for h

Using expressions for \( h \), equate them:\( x\sqrt{3} = \frac{y}{\sqrt{3}} \)Substitute \( y = h\sqrt{3} \) into \( x \cdot y = 1134 \):\( x \cdot h\sqrt{3} = 1134 \)Express \( x \) from \( h = x\sqrt{3} \):\( x = \frac{h}{\sqrt{3}} \)Substitute this into \( x \cdot h\sqrt{3} = 1134 \):\( \frac{h}{\sqrt{3}} \cdot h\sqrt{3} = 1134 \)\( h^2 = 1134 \)Solve for \( h \):\( h = \sqrt{1134} \approx 33.65 \)Rounding, the tree's height is closest to (A) 33.

Key Concepts

Angle of Elevation

The angle of elevation is a fundamental concept in trigonometry. It is the angle formed between the horizontal line from the observer's eye to a point higher than the observer. This concept is frequently used to determine heights or distances indirectly.

In the context provided, imagine standing on point B and looking up at the top of the tree. The "line of sight" would angle upwards at 60°. Similarly, moving to point C, the angle reduces to 30°, and further to D, it reduces to 15°. The decreasing angles of elevation as you move further away from the tree are tied directly to the geometry of this scenario and are critical in calculations involving the height of the tree.

Tangent Function

The tangent function is a cornerstone of trigonometry, particularly when dealing with right triangles. It relates the angle of a triangle to the ratio of the opposite side to the adjacent side. In simpler terms, for a given angle \( \theta \):

• \( \tan(\theta) = \frac{\text{opposite side}}{\text{adjacent side}} \)

By using the tangent function, you can solve for various unknowns when one side and an angle are known.

In our exercise, the height of the tree serves as the "opposite" side, and the distances from points B, C, and D to the tree act as the "adjacent" sides. The angles of elevation given for each point help set up these equations using tangent:

• \( \tan(60^{\circ}) = \sqrt{3} = \frac{h}{x} \) for point B,
• \( \tan(30^{\circ}) = \frac{1}{\sqrt{3}} = \frac{h}{y} \) for point C,
• \( \tan(15^{\circ}) = 2 - \sqrt{3} = \frac{h}{z} \) for point D.

Understanding how to manipulate these expressions is key to solving for the height.

Height of Tree

The ultimate goal in such problems often revolves around finding the height of some object, like a tree, without direct measurement.

Using our established angles of elevation and trigonometric functions, we can determine the height mathematically. By equating expressions derived from the tangent function and accommodating for relationships between the distances, you can solve for unknown variables.

In this exercise, the problem gives us the relationship between distances BC and CD which supports solving for the tree's height. With the previously derived equations:

• From point B, \( h = x \sqrt{3} \)
• From point C, \( h = \frac{y}{\sqrt{3}} \)
• Using the product \( x \cdot y = 1134 \), solutions lead us to: \( h^2 = 1134 \)

Thus, solving \( h = \sqrt{1134} \) approximates the height as 33, showcasing the technique's efficacy.