The angle of elevation of the top of a tower from a point $A$ due north of it is $\alpha$ and from a point $B$ at a distance of $9$ units due west of $A$ is $\cos^{-1}\left(\frac{3}{\sqrt{13}}\right)$. If the distance of the point $B$ from the tower is $15$ units, then $\cot \alpha$ is equal to : Options: (A) $\frac{6}{5}$ (B) $\frac{9}{5}$ (C) $\frac{4}{3}$ (D) $\frac{7}{3}$

### Analyzing the Setup Imagine a tower standing vertically on a flat plain. Let the base of the tower be $O$ and its top be $P$. We are given two observation points, $A$ and $B$, on the ground. We walk due North from $O$ to reach point $A$. Then, we turn West to reach point $B$. Since North and West are perpendicular, $\triangle OAB$ is a right-angled triangle with the right angle at $A$. ### The Ground Plane Geometry We are given the distances $AB = 9$ and $OB = 15$. According to the Pythagorean theorem in $\triangle OAB$: $$OA^2 + AB^2 = OB^2$$ Substituting the known values: $$OA^2 + 9^2 = 15^2$$ $$OA^2 + 81 = 225$$ $$OA^2 = 144$$ Thus, the horizontal distance from the tower base to point $A$ is $OA = 12$. ### The Vertical Ascent Let the height of the tower be $h$. From point $B$, the angle of elevation to the top of the tower $P$ is $\beta$, with $\cos \beta = \frac{3}{\sqrt{13}}$. In the vertical right-angled triangle $\triangle OBP$, the tangent of the angle of elevation is defined as: $$\tan \beta = \frac{h}{OB} = \frac{h}{15}$$ Using the trigonometric identity $\tan \beta = \frac{\sqrt{1 - \cos^2 \beta}}{\cos \beta}$, we calculate: $$\tan \beta = \frac{\sqrt{1 - \frac{9}{13}}}{\frac{3}{\sqrt{13}}} = \frac{\sqrt{\frac{4}{13}}}{\frac{3}{\sqrt{13}}} = \frac{2}{3}$$ Equating the two expressions for $\tan \beta$: $$\frac{h}{15} = \frac{2}{3} \Rightarrow h = 10$$ ### Final Calculation We now determine $\cot \alpha$, where $\alpha$ is the angle of elevation from point $A$. In the vertical triangle $\triangle OAP$: $$\tan \alpha = \frac{h}{OA} = \frac{10}{12} = \frac{5}{6}$$ Since $\cot \alpha$ is the reciprocal of $\tan \alpha$, we find: $$\cot \alpha = \frac{6}{5}$$ The final value is **$\frac{6}{5}$** (or **$1.2$**).

Why do we need to convert $\cos \beta$ to $\tan \beta$?

Think about the vertical triangle OBP. The angle of elevation $\beta$ is formed by the tower height (the opposite side) and the ground distance OB (the adjacent side). The tangent function is defined as $\frac{\text{opposite}}{\text{adjacent}}$, which directly links the height $h$ and the base $OB$. Since we are given $\cos \beta$, we use the identity $\tan \beta = \frac{\sqrt{1 - \cos^2 \beta}}{\cos \beta}$ or simply draw a reference triangle to find the tangent value.

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Animated Solution for Mathematics - Trigonometry: The angle of elevation of the top of a tower from a point $A$ due north of it is $α$ and from a point $B$ at a distance of $9$ units due west of $A$ is $cos^{- 1} (\frac{3}{13})$ . If the distance of the point $B$ from the tower is $15$ units, then $cot α$ is equal to :

Select Answer:

Visualized Solution

Visualizing the 3D Setup

Let the tower be $O P$ with height $h$ , where $O$ is the base on the ground.
Point $A$ is due North of $O$ , and point $B$ is due West of $A$ .

The Right-Angled Triangle on the Ground

Since North and West are perpendicular, $∠ O A B = 9 0^{\circ}$ .
We are given $A B = 9$ units and $O B = 15$ units.

Applying Pythagoras in $△ O A B$

In right $△ O A B$ , apply Pythagoras theorem: $O A^{2} + A B^{2} = O B^{2}$
Substitute the known values: $O A^{2} + 9^{2} = 1 5^{2}$

Calculating Distance $O A$

$O A^{2} + 81 = 225$
$O A^{2} = 225 - 81 = 144$
$O A = 12$ units

Angle of Elevation from Point $B$

Let $β$ be the angle of elevation of the top of the tower from $B$ .
Given: $cos β = \frac{3}{13}$

Converting $cos β$ to $tan β$

We need $tan β$ to relate height $h$ and base $O B$ .
$tan β = \frac{( 13 ) ^{2} - 3 ^{2}}{3}$
$tan β = \frac{13 - 9}{3} = \frac{4}{3} = \frac{2}{3}$

Finding the Tower Height $h$

In right $△ O B P$ , $tan β = \frac{O P}{O B} = \frac{h}{15}$
Equating the values: $\frac{h}{15} = \frac{2}{3}$
$h = \frac{2 \times 15}{3} = 10$ units

Angle of Elevation from Point $A$

The angle of elevation from point $A$ is given as $α$ .
In right $△ O A P$ , $tan α = \frac{O P}{O A}$

Calculating $tan α$

Substitute $h = 10$ and $O A = 12$ .
$tan α = \frac{10}{12}$
Simplifying the fraction: $tan α = \frac{5}{6}$

Final Conclusion for $cot α$

The question asks for $cot α$ .
$cot α = \frac{1}{t a n α}$
$cot α = \frac{6}{5}$

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The Sigma Insight: Heights and Distances

Solution Diagram

Analyzing the Setup

Imagine a tower standing vertically on a flat plain. Let the base of the tower be

O

and its top be

P

. We are given two observation points,

A

and

B

, on the ground.

We walk due North from

O

to reach point

A

. Then, we turn West to reach point

B

. Since North and West are perpendicular,

△ O A B

is a right-angled triangle with the right angle at

A

The Ground Plane Geometry

We are given the distances

A B = 9

and

O B = 15

. According to the Pythagorean theorem in

△ O A B

O A^{2} + A B^{2} = O B^{2}

Substituting the known values:

O A^{2} + 9^{2} = 1 5^{2}

O A^{2} + 81 = 225

O A^{2} = 144

Thus, the horizontal distance from the tower base to point

A

O A = 12

The Vertical Ascent

Let the height of the tower be

h

. From point

B

, the angle of elevation to the top of the tower

P

β

, with

cos β = \frac{3}{13}

In the vertical right-angled triangle

△ O B P

, the tangent of the angle of elevation is defined as:

tan β = \frac{h}{O B} = \frac{h}{15}

Using the trigonometric identity

tan β = \frac{1 - c o s ^{2} β}{c o s β}

, we calculate:

tan β = \frac{1 - \frac{9}{13}}{\frac{3}{13}} = \frac{\frac{4}{13}}{\frac{3}{13}} = \frac{2}{3}

Equating the two expressions for

tan β

\frac{h}{15} = \frac{2}{3} \Rightarrow h = 10

Final Calculation

We now determine

cot α

, where

α

is the angle of elevation from point

A

. In the vertical triangle

△ O A P

tan α = \frac{h}{O A} = \frac{10}{12} = \frac{5}{6}

Since

cot α

is the reciprocal of

tan α

, we find:

cot α = \frac{6}{5}

The final value is $\frac{6}{5}$ (or $1.2$ ).

Common Traps & FAQs

Wait, why is the triangle OAB a right-angled triangle? I thought the tower was vertical.

That is a great observation! The tower is indeed vertical, but the triangle OAB exists entirely on the horizontal ground plane. Because the problem specifies that point A is due North of the base O, and point B is due West of A, the directions North and West are perpendicular to each other. Therefore, the angle at A is

9 0^{\circ}

, forming a right-angled triangle on the ground.

Can I use the Law of Sines instead of the Pythagorean theorem?

You absolutely could, but it would be like using a sledgehammer to crack a nut! Since we are dealing with a right-angled triangle on the ground, the Pythagorean theorem is the most direct and efficient path. It keeps the algebra clean and minimizes the chance of calculation errors.

Why do we need to convert

cos β

tan β

Think about the vertical triangle OBP. The angle of elevation

β

is formed by the tower height (the opposite side) and the ground distance OB (the adjacent side). The tangent function is defined as

\frac{opposite}{adjacent}

, which directly links the height

h

and the base

O B

. Since we are given

cos β

, we use the identity

tan β = \frac{1 - c o s ^{2} β}{c o s β}

or simply draw a reference triangle to find the tangent value.

Select Answer:

Visualized Solution

Visualizing the 3D Setup

The Right-Angled Triangle on the Ground

Applying Pythagoras in $△ O A B$

Calculating Distance $O A$

Angle of Elevation from Point $B$

Converting $cos β$ to $tan β$

Finding the Tower Height $h$

Angle of Elevation from Point $A$

Calculating $tan α$

Final Conclusion for $cot α$

The Sigma Insight: Heights and Distances

Analyzing the Setup

The Ground Plane Geometry

The Vertical Ascent

Final Calculation

Similar Questions

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 :

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

Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P be a point on the ground such that AP = 2AB. If $∠ B P C = β$ , then $tan β$ is equal to:

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:

Select Answer:

Visualized Solution

Visualizing the 3D Setup

The Right-Angled Triangle on the Ground

Applying Pythagoras in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △OAB

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

Angle of Elevation from Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Converting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cosβ to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tanβ

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

Angle of Elevation from Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

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

Final Conclusion for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cotα

The Sigma Insight: Heights and Distances

Analyzing the Setup

The Ground Plane Geometry

The Vertical Ascent

Final Calculation

Similar Questions

Applying Pythagoras in $△ O A B$

Calculating Distance $O A$

Angle of Elevation from Point $B$

Converting $cos β$ to $tan β$

Finding the Tower Height $h$

Angle of Elevation from Point $A$

Calculating $tan α$

Final Conclusion for $cot α$