Why do we use $\cot \theta$ instead of $\tan \theta$? Isn't $\tan$ more common?

Great question! While $\tan \theta$ is the standard, using $\cot \theta$ is a strategic choice here. Since we want the base lengths ($AQ, BQ, CQ$) and we know the height ($h$), $\cot \theta = \frac{\text{Base}}{\text{Height}}$ gives us the base directly as $h \cot \theta$. If you used $\tan$, you would have to write $\text{Base} = \frac{h}{\tan \theta}$, which is mathematically identical but adds an extra step of division. In JEE, saving those few seconds and reducing algebraic clutter is how you win!

JEE Main 2015

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Animated Solution for Mathematics - Trigonometry: If the angles of elevation of the top of a tower from three collinear points $A, B$ and $C$ , on a line leading to the foot of the tower, are $3 0^{\circ}, 4 5^{\circ}$ and $6 0^{\circ}$ respectively, then the ratio, $A B : B C$ , is:

Select Answer:

Progressive Hints

Hint 1

Imagine the tower as a vertical pillar of height

h

. You have three observers at points

A

B

, and

C

on the ground. They are all looking up at the same top point

P

. The key is to visualize three right-angled triangles sharing the same height

h

but having different base lengths along the line leading to the tower's foot.

Hint 2

Instead of using the tangent function, which would force you to work with fractions like

\frac{h}{A Q}

, try using the cotangent function:

cot θ = \frac{Base}{Perpendicular}

. This allows you to express the base distances

A Q

B Q

, and

C Q

directly in terms of

h

and the angle of elevation.

Hint 3

Start by writing the expressions for the bases:

A Q = h cot 3 0^{\circ}

B Q = h cot 4 5^{\circ}

, and

C Q = h cot 6 0^{\circ}

. Once you have these, remember that the segment

A B = A Q - B Q

and

B C = B Q - C Q

. The ratio

\frac{A B}{B C}

will then simplify beautifully as the height

h

cancels out.

Visualized Solution

Visualizing the Tower $P Q$

Let the tower be $P Q$ with height $h$ .
Points $A, B, C$ are collinear on a line leading to the foot $Q$ .

Angles of Elevation

Angles of elevation from $A, B, C$ to the top $P$ are $3 0^{\circ}, 4 5^{\circ}, 6 0^{\circ}$ respectively.
$∠ P A Q = 3 0^{\circ}$ , $∠ P B Q = 4 5^{\circ}$ , $∠ P C Q = 6 0^{\circ}$ .

The Cotangent Relation

In a right-angled triangle, $cot θ = \frac{Base}{Perpendicular}$ .
For $△ P QX$ , the base distance is $X Q = h cot θ$ .

Calculating Base $A Q$

In $△ P A Q$ , base $A Q = h cot 3 0^{\circ}$ .
Since $cot 3 0^{\circ} = 3$ , we get $A Q = h 3$ .

Calculating Base $B Q$

In $△ P B Q$ , base $B Q = h cot 4 5^{\circ}$ .
Since $cot 4 5^{\circ} = 1$ , we get $B Q = h$ .

Calculating Base $C Q$

In $△ P C Q$ , base $C Q = h cot 6 0^{\circ}$ .
Since $cot 6 0^{\circ} = \frac{1}{3}$ , we get $C Q = \frac{h}{3}$ .

Finding Length $A B$

The distance $A B$ is the difference between $A Q$ and $B Q$ .
$A B = A Q - B Q = h 3 - h$ .
Factoring out $h$ : $A B = h (3 - 1)$ .

Finding Length $B C$

The distance $B C$ is the difference between $B Q$ and $C Q$ .
$B C = B Q - C Q = h - \frac{h}{3}$ .
Taking LCM: $B C = h (\frac{3 - 1}{3})$ .

Setting up the Ratio $\frac{A B}{B C}$

We need to find the ratio $\frac{A B}{B C}$ .
Substitute the expressions: $\frac{A B}{B C} = \frac{h ( 3 - 1 )}{h ( \frac{3 - 1}{3} )}$ .

Simplifying the Ratio

Cancel the common terms $h$ and $(3 - 1)$ from numerator and denominator.
$\frac{A B}{B C} = \frac{1}{\frac{1}{3}} = 3$ .

Final Conclusion

The required ratio $A B : B C$ is $3 : 1$ .
Key Takeaway: The unknown height $h$ cancels out, which is a common pattern in such problems.

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The Sigma Insight: Heights and Distances

Solution Diagram

Analyzing the Setup

Imagine a tower with top

P

and base

Q

standing on a flat plain. Three observers

A

B

, and

C

are positioned on a straight line leading to

Q

The angles of elevation to the top of the tower

P

are given as: From $A$ : $∠ P A Q = 3 0^{\circ}$ From

B

∠ P B Q = 4 5^{\circ}

* From

C

∠ P C Q = 6 0^{\circ}

Our objective is to determine the ratio of the distances

A B

B C

The Power of the Cotangent

Let the height of the tower be

h

. We consider the three right-angled triangles

△ P A Q

△ P B Q

, and

△ P C Q

Using the cotangent function, where

cot θ = \frac{Base}{Perpendicular}

, we express the base of each triangle as

h cot θ

* For

△ P A Q

A Q = h cot 3 0^{\circ} = h 3

* For

△ P B Q

B Q = h cot 4 5^{\circ} = h (1) = h

* For

△ P C Q

C Q = h cot 6 0^{\circ} = h (\frac{1}{3}) = \frac{h}{3}

The Algebra of Distances

We now calculate the lengths of the segments

A B

and

B C

by finding the differences between the ground distances:

A B = A Q - B Q = h 3 - h = h (3 - 1)

B C = B Q - C Q = h - \frac{h}{3} = h (1 - \frac{1}{3}) = h (\frac{3 - 1}{3})

The Elegant Cancellation

To find the ratio

\frac{A B}{B C}

, we substitute the expressions derived above:

\frac{A B}{B C} = \frac{h ( 3 - 1 )}{h ( \frac{3 - 1}{3} )}

Observe that the height

h

and the term

(3 - 1)

appear in both the numerator and the denominator. Canceling these common terms yields:

\frac{A B}{B C} = \frac{1}{\frac{1}{3}} = 3

The final ratio

A B : B C

is $3 : 1$ .

Common Traps & FAQs

Why do we use

cot θ

instead of

tan θ

? Isn't

tan

more common?

Great question! While

tan θ

is the standard, using

cot θ

is a strategic choice here. Since we want the base lengths (

A Q, B Q, C Q

) and we know the height (

h

cot θ = \frac{Base}{Height}

gives us the base directly as

h cot θ

. If you used

tan

, you would have to write

Base = \frac{h}{t a n θ}

, which is mathematically identical but adds an extra step of division. In JEE, saving those few seconds and reducing algebraic clutter is how you win!

Wait, the problem doesn't give the height of the tower

h

. How can we find the ratio without it?

That is the beauty of this problem! In many geometry problems involving ratios, the absolute scale (the actual height) doesn't matter—only the relative proportions do. When you set up the ratio

\frac{A B}{B C}

, the variable

h

appears in both the numerator and the denominator. Because it is a common factor, it cancels out completely. It is like asking for the ratio of two shadows; the height of the object casting them does not change the ratio of the shadows themselves.

Can I use

sin

cos

to solve this?

Technically, yes, but it would be like taking a long, winding road to reach a destination that is right in front of you. Using

sin

cos

would involve the hypotenuse (

A P, B P, C P

), which we don't know and don't need. You would end up with more variables and more complex equations. Always look for the trigonometric ratio that links the sides you know (or want to know) with the side that is constant—in this case, the height

h

Animated Solution for Mathematics - Trigonometry: If the angles of elevation of the top of a tower from three collinear points $A, B$ and $C$ , on a line leading to the foot of the tower, are $3 0^{\circ}, 4 5^{\circ}$ and $6 0^{\circ}$ respectively, then the ratio, $A B : B C$ , is:

Select Answer:

Visualized Solution

Visualizing the Tower $P Q$

Angles of Elevation

The Cotangent Relation

Calculating Base $A Q$

Calculating Base $B Q$

Calculating Base $C Q$

Finding Length $A B$

Finding Length $B C$

Setting up the Ratio $\frac{A B}{B C}$

Simplifying the Ratio

Final Conclusion

The Sigma Insight: Heights and Distances

Analyzing the Setup

The Power of the Cotangent

The Algebra of Distances

The Elegant Cancellation

Similar Questions

A tower stands at the centre of a circular park. $A$ and $B$ are two points on the boundary of the park such that $A B (= a)$ subtends an angle of $6 0^{\circ}$ at the foot of the tower, and the angle of elevation of the top of the tower from $A$ or $B$ is $3 0^{\circ}$ . The height of the tower is

ABC is a triangular park with AB = AC = 100 metres. A vertical tower is situated at the mid-point of BC. If the angles of elevation of the top of the tower at A and B are $cot^{- 1} (32)$ and $csc^{- 1} (22)$ respectively, then the height of the tower (in metres) is :

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:

$A B C$ is a triangular park with $A B = A C = 100$ m. A television tower stands at the midpoint of $B C$ . The angles of elevation of the top of the tower at $A, B, C$ are $4 5^{\circ}, 6 0^{\circ}, 6 0^{\circ}$ , respectively. Find the height of the tower.

Select Answer:

Visualized Solution

Visualizing the Tower .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PQ

Angles of Elevation

The Cotangent Relation

Calculating Base .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AQ

Calculating Base .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } BQ

Calculating Base .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CQ

Finding Length .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB

Finding Length .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } BC

Setting up the Ratio .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } BCAB​

Simplifying the Ratio

Final Conclusion

The Sigma Insight: Heights and Distances

Analyzing the Setup

The Power of the Cotangent

The Algebra of Distances

The Elegant Cancellation

Similar Questions

Visualizing the Tower $P Q$

Calculating Base $A Q$

Calculating Base $B Q$

Calculating Base $C Q$

Finding Length $A B$

Finding Length $B C$

Setting up the Ratio $\frac{A B}{B C}$