Two observers are stationed due north of a tower (of height x meter) at a distance y meter from each other. The angles of elevation of the tower observed by them are 30° and 45° respectively. Then x/y is equal to

Question

Prepp Team · Accepted Answer

This problem involves trigonometry, specifically using the tangent function, to relate the height of a tower to the distances of observers based on their angles of elevation.

Understanding the Tower Height and Angles of Elevation Problem

We are given a tower of height $x$ meters. Two observers are located due north of the tower at a distance $y$ meters from each other. They observe the top of the tower with angles of elevation $30^\circ$ and $45^\circ$. Since the observers are due north of the tower's base, they lie on a straight line extending north from the base.

Let the base of the tower be point A. Let the two observers be at points B and C. Since the angle of elevation is smaller for the observer further away, let the observer at C have the angle of elevation $30^\circ$, and the observer at B have the angle of elevation $45^\circ$. Points A, B, and C are collinear, with A, then B, then C (or A, then C, then B). Since both observers are due north, they are on a line North of the tower. The distance between them is given as $y$. The distance of the observer with the smaller angle ($30^\circ$) from the tower base will be greater than the distance of the observer with the larger angle ($45^\circ$).

Let the distance from the base of the tower (A) to the observer at B be $d_1$.

Let the distance from the base of the tower (A) to the observer at C be $d_2$.

The height of the tower is $x$.

The angles of elevation are $45^\circ$ at B and $30^\circ$ at C.

Since the observer with the $30^\circ$ angle is further away, $d_2 > d_1$. The distance between the observers is $y = d_2 - d_1$.

Applying Trigonometry with Angles of Elevation

We can form two right-angled triangles: one with the tower and observer B, and another with the tower and observer C. In a right-angled triangle, the tangent of an angle is the ratio of the side opposite the angle to the side adjacent to the angle.

For the observer at B (angle of elevation $45^\circ$):

The opposite side is the height of the tower ($x$).

The adjacent side is the distance from the tower base to B ($d_1$).

Using the tangent ratio:

$\tan(45^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{x}{d_1}$

We know that $\tan(45^\circ) = 1$. So,

$1 = \frac{x}{d_1}$

This gives us a relationship between $x$ and $d_1$:

$d_1 = x$

For the observer at C (angle of elevation $30^\circ$):

The opposite side is the height of the tower ($x$).

The adjacent side is the distance from the tower base to C ($d_2$).

Using the tangent ratio:

$\tan(30^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{x}{d_2}$

We know that $\tan(30^\circ) = \frac{1}{\sqrt{3}}$. So,

$\frac{1}{\sqrt{3}} = \frac{x}{d_2}$

This gives us a relationship between $x$ and $d_2$:

$d_2 = x\sqrt{3}$

Calculating the Ratio $x/y$

We are given that the distance between the two observers is $y$. Since they are on the same line due north of the tower, the distance $y$ is the difference between their distances from the base of the tower. As $d_2 > d_1$ (because the angle $30^\circ$ is smaller than $45^\circ$, meaning the observer is further away), the distance is:

$y = d_2 - d_1$

Substitute the values we found for $d_1$ and $d_2$ in terms of $x$:

$y = x\sqrt{3} - x$

We can factor out $x$ from the right side:

$y = x(\sqrt{3} - 1)$

The question asks for the ratio $\frac{x}{y}$. We can rearrange the equation above to find this ratio:

$\frac{x}{y} = \frac{1}{\sqrt{3} - 1}$

To simplify this expression and remove the square root from the denominator, we need to rationalize the denominator. We do this by multiplying both the numerator and the denominator by the conjugate of the denominator, which is $\sqrt{3} + 1$:

$\frac{x}{y} = \frac{1}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1}$

Multiply the numerators and the denominators:

Numerator: $1 \times (\sqrt{3} + 1) = \sqrt{3} + 1$

Denominator: $(\sqrt{3} - 1)(\sqrt{3} + 1)$. This is in the form $(a-b)(a+b) = a^2 - b^2$, where $a = \sqrt{3}$ and $b = 1$.

Denominator: $(\sqrt{3})^2 - (1)^2 = 3 - 1 = 2$

So, the ratio $\frac{x}{y}$ is:

$\frac{x}{y} = \frac{\sqrt{3} + 1}{2}$

This is the required value of the ratio $\frac{x}{y}$.

Revision Table: Key Trigonometry Values

Understanding standard trigonometric ratios for common angles is crucial for solving height and distance problems.

Angle	$\sin(\text{Angle})$	$\cos(\text{Angle})$	$\tan(\text{Angle})$
$0^\circ$	$0$	$1$	$0$
$30^\circ$	$\frac{1}{2}$	$\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{3}}$
$45^\circ$	$\frac{1}{\sqrt{2}}$	$\frac{1}{\sqrt{2}}$	$1$
$60^\circ$	$\frac{\sqrt{3}}{2}$	$\frac{1}{2}$	$\sqrt{3}$
$90^\circ$	$1$	$0$	Undefined

Additional Information on Height and Distance Problems

Height and distance problems are a classic application of trigonometry. They typically involve using angles of elevation or depression along with trigonometric ratios (sine, cosine, tangent) to find unknown heights or distances.

Angle of Elevation: When an observer looks upwards at an object, the angle between the horizontal line of sight and the line of sight to the object is called the angle of elevation.

Angle of Depression: When an observer looks downwards at an object, the angle between the horizontal line of sight and the line of sight to the object is called the angle of depression.

Solving these problems usually involves:

Drawing a clear diagram representing the problem.
Identifying the right triangles formed in the diagram.
Using the given angles and lengths, and the appropriate trigonometric ratio (SOH CAH TOA: Sine = Opposite/Hypotenuse, Cosine = Adjacent/Hypotenuse, Tangent = Opposite/Adjacent) to set up equations.
Solving the equations to find the unknown quantities.

These problems reinforce the practical application of trigonometry in real-world scenarios.

Prepp Team · Answer

$\frac{\sqrt{2}&minus;1}{2}$

Prepp Team · Answer

$\frac{\sqrt{3}&minus;1}{2}$

Prepp Team · Answer

1

Angle	\(\sin(\text{Angle})\)	\(\cos(\text{Angle})\)	\(\tan(\text{Angle})\)
\(0^\circ\)	\(0\)	\(1\)	\(0\)
\(30^\circ\)	\(\frac{1}{2}\)	\(\frac{\sqrt{3}}{2}\)	\(\frac{1}{\sqrt{3}}\)
\(45^\circ\)	\(\frac{1}{\sqrt{2}}\)	\(\frac{1}{\sqrt{2}}\)	\(1\)
\(60^\circ\)	\(\frac{\sqrt{3}}{2}\)	\(\frac{1}{2}\)	\(\sqrt{3}\)
\(90^\circ\)	\(1\)	\(0\)	Undefined

Understanding the Tower Height and Angles of Elevation Problem

Applying Trigonometry with Angles of Elevation

Calculating the Ratio \(x/y\)

Revision Table: Key Trigonometry Values

Additional Information on Height and Distance Problems

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