$ABC$ is an acute angled isosceles triangle. Two equal sides $AB$ and $AC$ lie on the lines $7x - y - 3 = 0$ and $x + y - 5 = 0$. If $\theta$ is one of the equal angles, then what is $\cot\theta$ equal to ?

Question

Prepp Team · Accepted Answer

Understanding the Problem: Isosceles Triangle Geometry

We are given an acute angled isosceles triangle, named $ABC$. The problem states that two sides of equal length, $AB$ and $AC$, lie along specific straight lines. We need to find the value of the cotangent of one of the equal angles, denoted by $\theta$. The lines containing sides $AB$ and $AC$ are given by the equations:

Line 1 ($AB$): $7x - y - 3 = 0$
Line 2 ($AC$): $x + y - 5 = 0$

In an isosceles triangle $ABC$ where $AB = AC$, the angles opposite to these sides are equal. These are $\angle B$ and $\angle C$. Therefore, the problem defines $\theta = \angle B = \angle C$. The angle $\angle A$ is the angle between the two given lines representing sides $AB$ and $AC$. The triangle is specified as acute-angled, meaning all its internal angles are less than $90^\circ$.

Step-by-Step Solution

Step 1: Find the angle $\angle A$ between the given lines

The vertex $A$ of the triangle is the intersection point of the lines $7x - y - 3 = 0$ and $x + y - 5 = 0$. Let $\phi$ represent the angle $\angle A$ between these two lines.

First, we determine the slopes of the lines:

For the line $7x - y - 3 = 0$, rearranging to the form $y = mx + c$, we get $y = 7x - 3$. The slope is $m_1 = 7$.
For the line $x + y - 5 = 0$, rearranging gives $y = -x + 5$. The slope is $m_2 = -1$.

The tangent of the angle $\phi$ between two lines with slopes $m_1$ and $m_2$ can be calculated using the formula:

$\tan\phi = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$

Substitute the slopes $m_1 = 7$ and $m_2 = -1$ into the formula:

$\tan\phi = \left| \frac{7 - (-1)}{1 + (7)(-1)} \right| = \left| \frac{7 + 1}{1 - 7} \right| = \left| \frac{8}{-6} \right| = \left| -\frac{4}{3} \right| = \frac{4}{3}$

So, the tangent of angle $A$ is $\tan A = \tan\phi = \frac{4}{3}$.

Step 2: Relate $\angle A$ and the equal angles $\theta$

The sum of the interior angles in any triangle is always $180^\circ$. For our triangle $ABC$, this means:

$\angle A + \angle B + \angle C = 180^\circ$

Given that the triangle is isosceles with $AB = AC$, the angles opposite these sides are equal: $\angle B = \angle C = \theta$. Substituting $\angle A = \phi$ and $\angle B = \angle C = \theta$ into the sum equation:

$\phi + \theta + \theta = 180^\circ$ $\phi + 2\theta = 180^\circ$

Step 3: Express $\cot\theta$ in terms of $\angle A$

Our goal is to find $\cot\theta$. We can use the relationship derived in Step 2:

$\phi + 2\theta = 180^\circ$

Rearranging the equation to solve for $\theta$:

$2\theta = 180^\circ - \phi$ $\theta = 90^\circ - \frac{\phi}{2}$

Now, we can find the cotangent of $\theta$:

$\cot\theta = \cot\left(90^\circ - \frac{\phi}{2}\right)$

Using the trigonometric identity $\cot(90^\circ - x) = \tan(x)$, where $x = \frac{\phi}{2}$, we get:

$\cot\theta = \tan\left(\frac{\phi}{2}\right)$

This means finding $\cot\theta$ is equivalent to finding $\tan(A/2)$, where $\tan A = 4/3$. We need to calculate the tangent of half the angle $A$.

Step 4: Calculate $\tan(A/2)$ using the half-angle formula

We use the tangent half-angle identity, which connects the tangent of an angle to the tangent of half that angle:

$\tan\phi = \frac{2\tan(\phi/2)}{1 - \tan^2(\phi/2)}$

Let $t = \tan(\phi/2)$. We know $\tan\phi = 4/3$. Substituting this into the identity:

$\frac{4}{3} = \frac{2t}{1 - t^2}$

To solve for $t$, we first cross-multiply:

$4(1 - t^2) = 3(2t)$ $4 - 4t^2 = 6t$

Rearrange this into the standard form of a quadratic equation ($ax^2 + bx + c = 0$):

$4t^2 + 6t - 4 = 0$

Simplify the equation by dividing all terms by 2:

$2t^2 + 3t - 2 = 0$

Factor the quadratic expression:

$(2t - 1)(t + 2) = 0$

This equation yields two possible solutions for $t$:

$2t - 1 = 0 \implies t = \frac{1}{2}$ $t + 2 = 0 \implies t = -2$

So, the possible values for $\tan(A/2)$ are $1/2$ and $-2$.

Step 5: Select the correct value using the acute angle condition

The problem specifies that triangle $ABC$ is an acute angled triangle. This implies that all its angles, including $\angle A = \phi$, must be less than $90^\circ$.

Since $0^\circ < \phi < 90^\circ$, it follows that $0^\circ < \phi/2 < 45^\circ$.

The tangent function is positive for angles between $0^\circ$ and $45^\circ$, and its value is less than 1. Therefore, we must have $0 < \tan(\phi/2) < 1$.

Let's examine our two possible values for $t = \tan(\phi/2)$:

$t = 1/2$. This value satisfies the condition $0 < t < 1$.
$t = -2$. This value is negative, so it does not satisfy the condition for an acute angle $A/2$.

Thus, the only valid value for $\tan(A/2)$ is $1/2$.

Since we established that $\cot\theta = \tan(A/2)$, we conclude:

$\cot\theta = \frac{1}{2}$

We can quickly check if this result is consistent with the acute angle condition. If $\cot\theta = 1/2$, then $\tan\theta = 2$. Since $\tan A = 4/3$ and $\tan\theta = 2$, both angle $A$ (approx $53.1^\circ$) and angle $\theta$ (approx $63.4^\circ$) are acute. The angles of the triangle are approximately $53.1^\circ, 63.4^\circ, 63.4^\circ$, all of which are less than $90^\circ$, confirming the triangle is acute-angled.

Final Answer

The value of $\cot\theta$ for the given acute angled isosceles triangle is $\frac{1}{2}$.

Prepp Team · Answer

$\frac{1}{3}$

Prepp Team · Answer

$\frac{2}{3}$

Prepp Team · Answer

2

Understanding the Problem: Isosceles Triangle Geometry

Step-by-Step Solution

Step 1: Find the angle \(\angle A\) between the given lines

Step 2: Relate \(\angle A\) and the equal angles \(\theta\)

Step 3: Express \(\cot\theta\) in terms of \(\angle A\)

Step 4: Calculate \(\tan(A/2)\) using the half-angle formula

Step 5: Select the correct value using the acute angle condition

Final Answer

Similar Questions

Important Questions from Equation in a Line

NDA 2026 Mock Test Series (Latest Pattern)

NDA Mock Test Series for Maths & GAT

Start Your Preparation with Prepp Mobile App