Question

How do you find the three angles of the triangle with the given vertices: A (1,0) B (4,6) C (-3,5)?

Answer 1

Hint : In this question, we are given three coordinates of a triangle and we have to find the angle between its three sides. Joining any two points, we will get one side of the triangle. A triangle is a two-dimensional figure so it lies in the XY plane. By finding the slope of the three sides of the triangle, we can easily find out the angle between the three sides.

Complete step-by-step answer :
The slope of the line joining the points A (1,0) and B (4,6) is ${m_1} = \dfrac{{6 - 0}}{{4 - 1}} = \dfrac{6}{3} = 2$
The slope of the line joining the points B (4,6) and C (-3,5) is ${m_2} = \dfrac{{5 - 6}}{{ - 3 - 4}} = \dfrac{{ - 1}}{{ - 7}} = \dfrac{1}{7}$
The slope of the line joining the points A (1,0) and C (-3,5) is ${m_3} = \dfrac{{5 - 0}}{{ - 3 - 1}} = - \dfrac{5}{4}$
The formula for finding the angle between two lines is – $\tan \theta = \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|$
So, the angle between the line AB and BC is –
$\tan {\theta _1} = \left| {\dfrac{{2 - \dfrac{1}{7}}}{{1 + 2 \times \dfrac{1}{7}}}} \right| \\\ \Rightarrow \tan {\theta _1} = \left| {\dfrac{{\dfrac{{13}}{7}}}{{\dfrac{{7 + 2}}{7}}}} \right| = \dfrac{{13}}{7} \times \dfrac{7}{9} \\\ \Rightarrow \tan {\theta _1} = \dfrac{{13}}{9} \\\ \Rightarrow {\theta _1} = {\tan ^{ - 1}}\dfrac{{13}}{9} = 55.3^\circ \;$
The angle between BC and AC is –
$\tan {\theta _2} = \left| {\dfrac{{\dfrac{1}{7} - ( - \dfrac{5}{4})}}{{1 + \dfrac{1}{7} \times ( - \dfrac{5}{4})}}} \right| \\\ \Rightarrow \tan {\theta _2} = \left| {\dfrac{{\dfrac{{4 + 35}}{{28}}}}{{\dfrac{{28 - 5}}{{28}}}}} \right| = \dfrac{{39}}{{28}} \times \dfrac{{28}}{{23}} \\\ \Rightarrow \tan {\theta _2} = \dfrac{{39}}{{23}} \\\ \Rightarrow {\theta _2} = {\tan ^{ - 1}}\dfrac{{39}}{{23}} = 59.5^\circ \;$
The angle between AB and AC is –
$\tan {\theta _3} = \left| {\dfrac{{2 - ( - \dfrac{5}{4})}}{{1 + 2 \times ( - \dfrac{5}{4})}}} \right| \\\ \Rightarrow \tan {\theta _3} = \left| {\dfrac{{\dfrac{{8 + 5}}{4}}}{{\dfrac{{4 - 10}}{4}}}} \right| = \dfrac{{13}}{4} \times \dfrac{4}{{ - 6}} \\\ \Rightarrow \tan {\theta _3} = \dfrac{{13}}{6} \\\ \Rightarrow {\theta _3} = {\tan ^{ - 1}}( \dfrac{{13}}{6}) = 65.2^\circ \;$ .
Hence, the three angles of the triangle are $55.3^\circ ,\,59.5^\circ \,\,and\,\,65.2^\circ$
So, the correct answer is “ $55.3^\circ ,\,59.5^\circ \,\,and\,\,65.2^\circ$ ”.

Note : When a line lies in the plane containing x and y axes, the change in the y-coordinate divided by the corresponding change in the x-coordinate between two distinct points of the line gives the slope of a line, that is, the slope of a line joining two points $({x_1},{y_1})$ and $({x_2},{y_2})$ is given by the formula $\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$ . Using this approach, similar questions can be solved easily. This question can also be solved using the laws of cosines