NIMCET 2010 — Mathematics PYQ

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Question

NIMCET 2010 — Mathematics PYQ

NIMCET | Mathematics | 2010

In a triangle $ABC$ , where $A = (-4, 2)$ , $B = (2, 1)$ and $C = (-4, -5)$ , find $\angle ACB$ .

Choose the correct answer:

Explanation

1. Slope of line $CA$ ( $m_1$ ):

Points: $C(-4, -5)$ and $A(-4, 2)$

m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - (-5)}{-4 - (-4)} = \frac{7}{0} = \infty

Since the slope is undefined, line $CA$ is a vertical line (parallel to the y-axis).

2. Slope of line $CB$ ( $m_2$ ):

Points: $C(-4, -5)$ and $B(2, 1)$

m_2 = \frac{1 - (-5)}{2 - (-4)} = \frac{6}{6} = 1

3. Angle Calculation:

The angle $\theta$ between a vertical line and a line with slope $m$ is given by:

\tan(90^\circ - \theta) = m \quad \text{or simply use } \tan \phi = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|

Since $CA$ is vertical ( $x = -4$ ) and the slope of $CB$ is $1$ :

The angle that $CB$ makes with the positive x-axis is $45^\circ$ (because $\tan 45^\circ = 1$ ).

A vertical line makes an angle of $90^\circ$ with the x-axis.

\angle ACB = |90^\circ - 45^\circ| = 45^\circ

Correct Option:

(b) 45°