Question

Equation of line BC is 2x+y-10=0. Point A is such that angle ABC = angle ACB = 30. The sum of the slopes of line AB and line AC is?

Note: Could you please use trigonometry in solving this question?

Answer 1

Answer on Question#51031 - Math - Trigonometry

Equation of line $BC$ is $2x + y - 10 = 0$ . Point $A$ is such that angle $ABC = \text{angle } ACB = \alpha = 30{}^\circ$ . The sum of the slopes of line $AB$ and line $AC$ is?

Note: Could you please use trigonometry in solving this question?

Solution:

Let's rewrite the equation of line $BC$ in the following way

$$y = 1 0 - 2 x$$

The slope of this line gives us the tangent of the angle which this line creates with x-axis. So

$$\tan \varphi = - 2$$

Line $AB$ creates the angle $\varphi - \alpha$ with the x-axis and its slope is

$$\tan (\varphi - \alpha) = \frac {\tan \varphi - \tan \alpha}{1 + \tan \varphi \cdot \tan \alpha} = \frac {- 2 - \frac {1}{\sqrt {3}}}{1 - \frac {2}{\sqrt {3}}} = \frac {2 \sqrt {3} + 1}{2 - \sqrt {3}}$$

Line $AC$ creates the angle $\varphi + \alpha$ with the x-axis and its slope is

$$\tan (\varphi + \alpha) = \frac {\tan \varphi + \tan \alpha}{1 - \tan \varphi \cdot \tan \alpha} = \frac {- 2 + \frac {1}{\sqrt {3}}}{1 + \frac {2}{\sqrt {3}}} = \frac {1 - 2 \sqrt {3}}{\sqrt {3} + 2}$$

The sum of slopes of lines $AB$ and $AC$ is

$$\begin{array}{l} \tan (\varphi - \alpha) + \tan (\varphi + \alpha) = \frac {2 \sqrt {3} + 1}{2 - \sqrt {3}} + \frac {1 - 2 \sqrt {3}}{2 + \sqrt {3}} = \\ = \frac {\left(2 \sqrt {3} + 1\right) \left(\sqrt {3} + 2\right) + \left(1 - 2 \sqrt {3}\right) \left(2 - \sqrt {3}\right)}{2 ^ {2} - \sqrt {3} ^ {2}} = 16 \end{array}$$

If we consider the case when the point $A$ lies under the line $BC$ the answer will be the same, since the angles which lines $AB$ and $AC$ create with $x$-axis will be $\varphi + \alpha$ and $\varphi - \alpha$ correspondingly. So the sum of slopes will be the same.

**Answer:** 16.

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