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Animated Solution for Mathematics - Trigonometry: The upper $\frac{3}{4}$ th portion of a vertical pole subtends an angle $tan^{- 1} \frac{3}{5}$ at a point in the horizontal plane through its foot and at a distance 40 m from the foot. A possible height of the vertical pole is

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Progressive Hints

Hint 1

Hint 1 (The Spark): Visualize the geometry. You have a vertical pole of height

h

and a point

P

on the ground

40 m

from the foot. The pole is divided into two segments: a lower part of height

\frac{1}{4} h

and an upper part of height

\frac{3}{4} h

. The upper part subtends an angle

β = tan^{- 1} \frac{3}{5}

P

. Can you see two right-angled triangles sharing the same base of

40 m

Hint 2

Hint 2 (The Arsenal): You need to relate the angle

β

to the angles subtended by the whole pole and the lower part. Let

α

be the angle subtended by the lower part. Then the angle subtended by the whole pole is

α + β

. Use the tangent subtraction formula:

tan β = tan ((α + β) - α) = \frac{t a n ( α + β ) - t a n α}{1 + t a n ( α + β ) t a n α}

Hint 3

Hint 3 (The Execution): Express the tangents in terms of

h

tan (α + β) = \frac{h}{40}

and

tan α = \frac{h /4}{40} = \frac{h}{160}

. Now, substitute these into the formula

\frac{3}{5} = \frac{\frac{h}{40} - \frac{h}{160}}{1 + ( \frac{h}{40} ) ( \frac{h}{160} )}

and solve the resulting quadratic equation for

h

Visualized Solution

Visualizing the Setup

Let the total height of the vertical pole be $h$ meters.
The observation point $P$ is at a distance of $40 m$ from the foot $F$ of the pole.

Defining the Segments

Lower portion height = $\frac{1}{4} h$
Upper portion height = $\frac{3}{4} h$

Defining the Angles $α$ and $β$

Let $α$ be the angle subtended by the lower portion at $P$ .
Let $β$ be the angle subtended by the upper portion at $P$ .
Given: $β = tan^{- 1} \frac{3}{5}$

Setting up Trigonometric Ratios

In the lower right triangle: $tan α = \frac{h /4}{40} = \frac{h}{160}$
In the full right triangle: $tan (α + β) = \frac{h}{40}$

The Compound Angle Formula

We can express $β$ as $(α + β) - α$ .
Using the identity: $tan β = tan ((α + β) - α)$
$tan β = \frac{t a n ( α + β ) - t a n α}{1 + t a n ( α + β ) t a n α}$

Substitution of Values

Substitute the known values into the formula:
$\frac{3}{5} = \frac{\frac{h}{40} - \frac{h}{160}}{1 + ( \frac{h}{40} ) ( \frac{h}{160} )}$

Simplifying the Numerator

Numerator: $\frac{h}{40} - \frac{h}{160}$
Common denominator is $160$ : $\frac{4 h - h}{160} = \frac{3 h}{160}$

Simplifying the Denominator

Denominator: $1 + (\frac{h}{40}) (\frac{h}{160})$
Multiply the terms: $1 + \frac{h ^{2}}{6400}$
Combine: $\frac{6400 + h ^{2}}{6400}$

Forming the Equation

Combine numerator and denominator:
$\frac{3}{5} = \frac{\frac{3 h}{160}}{\frac{6400 + h ^{2}}{6400}}$
$\frac{3}{5} = \frac{3 h \cdot 6400}{160 ( 6400 + h ^{2} )} = \frac{120 h}{6400 + h ^{2}}$

Cross-Multiplication

Cancel $3$ from both sides: $\frac{1}{5} = \frac{40 h}{6400 + h ^{2}}$
Cross-multiply: $6400 + h^{2} = 200 h$

The Quadratic Equation

Rearrange into standard form:
$h^{2} - 200 h + 6400 = 0$

Solving for Height $h$

Factorize the quadratic: $h^{2} - 160 h - 40 h + 6400 = 0$
$(h - 160) (h - 40) = 0$
Possible values: $h = 160 m$ or $h = 40 m$

Conclusion

Final Answer: The possible height of the pole is $40 m$ .
Key Concept: Using the compound angle formula $tan (A - B)$ to relate subtended angles.

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The Sigma Insight: Heights and Distances

Solution Diagram

Analyzing the Setup

Imagine you are a surveyor standing on a flat, horizontal plane, looking up at a tall, vertical pole. You are exactly

40 m

away from its base.

The pole is divided into two distinct segments. The upper three-fourths of the pole subtends a specific angle at your eye, given by

β = tan^{- 1} \frac{3}{5}

Let the total height of the pole be

h

. The pole is split into a lower segment of height

\frac{1}{4} h

and an upper segment of height

\frac{3}{4} h

The Trigonometric Bridge

To solve this, we connect the given angle

β

to the physical dimensions of the pole. Let

α

be the angle subtended by the lower segment at our observation point.

The angle subtended by the entire pole is then

α + β

. We can now form two right-angled triangles based on these angles.

For the smaller triangle (lower segment):

tan α = \frac{h /4}{40} = \frac{h}{160}

For the larger triangle (entire pole):

tan (α + β) = \frac{h}{40}

We utilize the compound angle formula for tangent to bridge these values:

tan β = tan ((α + β) - α) = \frac{tan ( α + β ) - tan α}{1 + tan ( α + β ) tan α}

The Algebraic Dance

Substituting our expressions into the formula, we obtain:

\frac{3}{5} = \frac{\frac{h}{40} - \frac{h}{160}}{1 + ( \frac{h}{40} ) ( \frac{h}{160} )}

Simplifying the numerator:

\frac{h}{40} - \frac{h}{160} = \frac{4 h - h}{160} = \frac{3 h}{160}

Simplifying the denominator:

1 + \frac{h ^{2}}{6400} = \frac{6400 + h ^{2}}{6400}

Combining these into the master equation:

\frac{3}{5} = \frac{3 h}{160} \cdot \frac{6400}{6400 + h ^{2}} = \frac{120 h}{6400 + h ^{2}}

Dividing both sides by

3

and cross-multiplying:

\frac{1}{5} = \frac{40 h}{6400 + h ^{2}} \Rightarrow 6400 + h^{2} = 200 h

The Final Revelation

Rearranging the expression into a standard quadratic equation:

h^{2} - 200 h + 6400 = 0

Factorizing the quadratic:

(h - 160) (h - 40) = 0

This yields two possible heights for the pole: $h = 160 m$ or $h = 40 m$

Common Traps & FAQs

Wait, why do we get two answers for

h

? Is the pole height ambiguous?

That is a fantastic observation! In geometry, the angle subtended by a segment at a point on the ground can be the same for two different heights. Imagine moving the observation point closer or further; the angle changes. Mathematically, the quadratic equation

h^{2} - 200 h + 6400 = 0

naturally yields two roots,

h = 40

and

h = 160

. Both are physically valid solutions for the given angle condition. In a multiple-choice context, you simply select the one that appears in the options.

Why can't we just use the sine rule instead of the compound angle formula?

You absolutely can! The sine rule is a powerful alternative. If you define the triangles and use the sine rule in the triangle formed by the observation point and the two ends of the upper segment, you will arrive at the same result. However, the compound angle formula is often faster and less prone to algebraic errors in this specific setup. Both methods are elegant; the key is to choose the one you are most comfortable with.

Is there a way to solve this without the compound angle formula?

Yes, you can use the geometric properties of the triangles directly. By expressing the lengths of the segments in terms of the angles and the distance, you can set up a system of equations. However, the compound angle formula is the most direct 'bridge' between the given angle

β

and the unknown height

h

. It is a classic JEE technique to convert trigonometric conditions into algebraic equations.

Animated Solution for Mathematics - Trigonometry: The upper $\frac{3}{4}$ th portion of a vertical pole subtends an angle $tan^{- 1} \frac{3}{5}$ at a point in the horizontal plane through its foot and at a distance 40 m from the foot. A possible height of the vertical pole is

Select Answer:

Visualized Solution

Visualizing the Setup

Defining the Segments

Defining the Angles $α$ and $β$

Setting up Trigonometric Ratios

The Compound Angle Formula

Substitution of Values

Simplifying the Numerator

Simplifying the Denominator

Forming the Equation

Cross-Multiplication

The Quadratic Equation

Solving for Height $h$

Conclusion

The Sigma Insight: Heights and Distances

Analyzing the Setup

The Trigonometric Bridge

The Algebraic Dance

The Final Revelation

Similar Questions

A vertical pole stands at a point $Q$ on a horizontal ground. $A$ and $B$ are points on the ground, $d$ meters apart. The pole subtends angles $α$ and $β$ at $A$ and $B$ respectively. $A B$ subtends an angle $γ$ at $Q$ . Find the height of the pole.

From the base of a pole of height 20 meter, the angle of elevation of the top of a tower is $6 0^{\circ}$ . The pole subtends an angle $3 0^{\circ}$ at the top of the tower. Then the height of the tower is:

Two vertical poles of heights, 20m and 80m stand a part on a horizontal plane. The height (in meters) of the point of intersection of the lines joining the top of each pole to the foot of the other, from this horizontal plane is :

A pole stands vertically inside a triangular park $A B C$ . Let the angle of elevation of the top of the pole from each corner of the park be $\frac{π}{3}$ . If the radius of the circumcircle of $Δ A B C$ is 2, then the height of the pole is equal to:

A vertical pole fixed to the horizontal ground is divided in the ratio $3 : 7$ by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a point on the ground $18 m$ away from the base of the pole, then the height of the pole (in meters) is :

Two vertical poles are 150 m apart and the height of one is three times that of the other. If from the middle point of the line joining their feet, an observer finds the angles of elevation of their tops to be complementary, then the height of the shorter pole (in meters) is:

Two vertical poles $A B = 15 m$ and $C D = 10 m$ are standing apart on a horizontal ground with points $A$ and $C$ on the ground. If $P$ is the point of intersection of $B C$ and $A D$ , then the height of $P$ (in $m$ ) above the line $A C$ is :

Select Answer:

Visualized Solution

Visualizing the Setup

Defining the Segments

Defining the Angles .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β

Setting up Trigonometric Ratios

The Compound Angle Formula

Substitution of Values

Simplifying the Numerator

Simplifying the Denominator

Forming the Equation

Cross-Multiplication

The Quadratic Equation

Solving for Height .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h

Conclusion

The Sigma Insight: Heights and Distances

Analyzing the Setup

The Trigonometric Bridge

The Algebraic Dance

The Final Revelation

Similar Questions

Two vertical poles of heights, 20m and 80m stand a part on a horizontal plane. The height (in meters) of the point of intersection of the lines joining the top of each pole to the foot of the other, from this horizontal plane is :

Two vertical poles are 150 m apart and the height of one is three times that of the other. If from the middle point of the line joining their feet, an observer finds the angles of elevation of their tops to be complementary, then the height of the shorter pole (in meters) is:

Defining the Angles $α$ and $β$

Solving for Height $h$