Question Statement Find the interior angles of the triangle whose vertices are:
A(-2, 11) B(-6, -3) C(4, -9)
Background and Explanation To find the interior angles of a triangle from its vertices, we need to:
Calculate the slopes of the sides using the formula for the slope between two points.
Use the formula for the tangent of the angle between two lines to compute the angles at each vertex.
The slope of a line joining two points ( x 1 , y 1 ) (x_1, y_1) ( x 1 β , y 1 β ) ( x 2 , y 2 ) (x_2, y_2) ( x 2 β , y 2 β ) m = y 2 β y 1 x 2 β x 1 m = \frac{y_2 - y_1}{x_2 - x_1} m = x 2 β β x 1 β y 2 β β y 1 β β
To find the angle between two lines with slopes m 1 m_1 m 1 β m 2 m_2 m 2 β tan β‘ ( ΞΈ ) = β£ m 1 β m 2 1 + m 1 m 2 β£ \tan(\theta) = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right| tan ( ΞΈ ) = β 1 + m 1 β m 2 β m 1 β β m 2 β β β
Solution Step 1: Find the Slopes
The slopes of the sides are calculated as follows:
Slope of AB:
m 1 = β 3 β 11 β 6 β ( β 2 ) = β 14 β 4 = 7 2 m_1 = \frac{-3 - 11}{-6 - (-2)} = \frac{-14}{-4} = \frac{7}{2} m 1 β = β 6 β ( β 2 ) β 3 β 11 β = β 4 β 14 β = 2 7 β
Slope of BC:
m 2 = β 9 β ( β 3 ) 4 β ( β 6 ) = β 6 10 = β 3 5 m_2 = \frac{-9 - (-3)}{4 - (-6)} = \frac{-6}{10} = -\frac{3}{5} m 2 β = 4 β ( β 6 ) β 9 β ( β 3 ) β = 10 β 6 β = β 5 3 β
Slope of AC:
m 3 = β 9 β 11 4 β ( β 2 ) = β 20 6 = β 10 3 m_3 = \frac{-9 - 11}{4 - (-2)} = \frac{-20}{6} = -\frac{10}{3} m 3 β = 4 β ( β 2 ) β 9 β 11 β = 6 β 20 β = β 3 10 β
Step 2: Calculate the Angles
Now we calculate the angles using the formula for the tangent of the angle between two lines.
Angle at vertex A:
The angle at vertex A is between the lines AB and AC:
tan β‘ ( β A ) = β£ m 1 β m 3 1 + m 1 m 3 β£ = β£ 7 2 β ( β 10 3 ) 1 + 7 2 β
( β 10 3 ) β£ \tan(\angle A) = \left| \frac{m_1 - m_3}{1 + m_1 m_3} \right| = \left| \frac{\frac{7}{2} - \left(-\frac{10}{3}\right)}{1 + \frac{7}{2} \cdot \left(-\frac{10}{3}\right)} \right| tan ( β A ) = β 1 + m 1 β m 3 β m 1 β β m 3 β β β = β 1 + 2 7 β β
( β 3 10 β ) 2 7 β β ( β 3 10 β ) β β
Simplifying:
tan β‘ ( β A ) = 10 3 + 7 2 1 β 35 6 = 41 6 1 6 = 41 64 \tan(\angle A) = \frac{\frac{10}{3} + \frac{7}{2}}{1 - \frac{35}{6}} = \frac{\frac{41}{6}}{\frac{1}{6}} = \frac{41}{64} tan ( β A ) = 1 β 6 35 β 3 10 β + 2 7 β β = 6 1 β 6 41 β β = 64 41 β
Now, calculate the inverse tangent:
β A = tan β‘ β 1 ( 41 64 ) β 32.6 4 β \angle A = \tan^{-1}\left(\frac{41}{64}\right) \approx 32.64^\circ β A = tan β 1 ( 64 41 β ) β 32.6 4 β
Angle at vertex B:
The angle at vertex B is between the lines AB and BC:
tan β‘ ( β B ) = β£ m 1 β m 2 1 + m 1 m 2 β£ = β£ 7 2 β ( β 3 5 ) 1 + 7 2 β
( β 3 5 ) β£ \tan(\angle B) = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{\frac{7}{2} - \left(-\frac{3}{5}\right)}{1 + \frac{7}{2} \cdot \left(-\frac{3}{5}\right)} \right| tan ( β B ) = β 1 + m 1 β m 2 β m 1 β β m 2 β β β = β 1 + 2 7 β β
( β 5 3 β ) 2 7 β β ( β 5 3 β ) β β
Simplifying:
tan β‘ ( β B ) = 7 2 + 3 5 1 β 21 10 = 35 10 + 6 10 β 11 10 = 41 11 \tan(\angle B) = \frac{\frac{7}{2} + \frac{3}{5}}{1 - \frac{21}{10}} = \frac{\frac{35}{10} + \frac{6}{10}}{-\frac{11}{10}} = \frac{41}{11} tan ( β B ) = 1 β 10 21 β 2 7 β + 5 3 β β = β 10 11 β 10 35 β + 10 6 β β = 11 41 β
Now, calculate the inverse tangent:
β B = tan β‘ β 1 ( 41 11 ) β 105.0 2 β \angle B = \tan^{-1}\left(\frac{41}{11}\right) \approx 105.02^\circ β B = tan β 1 ( 11 41 β ) β 105.0 2 β
Angle at vertex C:
The angle at vertex C is between the lines BC and AC:
tan β‘ ( β C ) = β£ m 2 β m 3 1 + m 2 m 3 β£ = β£ β 3 5 β ( β 10 3 ) 1 + ( β 3 5 ) β
( β 10 3 ) β£ \tan(\angle C) = \left| \frac{m_2 - m_3}{1 + m_2 m_3} \right| = \left| \frac{-\frac{3}{5} - \left(-\frac{10}{3}\right)}{1 + \left(-\frac{3}{5}\right) \cdot \left(-\frac{10}{3}\right)} \right| tan ( β C ) = β 1 + m 2 β m 3 β m 2 β β m 3 β β β = β 1 + ( β 5 3 β ) β
( β 3 10 β ) β 5 3 β β ( β 3 10 β ) β β
Simplifying:
tan β‘ ( β C ) = β 9 15 + 50 15 1 + 30 15 = 41 15 45 15 = 41 25 \tan(\angle C) = \frac{\frac{-9}{15} + \frac{50}{15}}{1 + \frac{30}{15}} = \frac{\frac{41}{15}}{\frac{45}{15}} = \frac{41}{25} tan ( β C ) = 1 + 15 30 β 15 β 9 β + 15 50 β β = 15 45 β 15 41 β β = 25 41 β
Now, calculate the inverse tangent:
β C = tan β‘ β 1 ( 41 25 ) β 42.3 4 β \angle C = \tan^{-1}\left(\frac{41}{25}\right) \approx 42.34^\circ β C = tan β 1 ( 25 41 β ) β 42.3 4 β
Slope Formula:
m = y 2 β y 1 x 2 β x 1 m = \frac{y_2 - y_1}{x_2 - x_1} m = x 2 β β x 1 β y 2 β β y 1 β β
Tangent Formula for Angle Between Two Lines:
tan β‘ ( ΞΈ ) = β£ m 1 β m 2 1 + m 1 m 2 β£ \tan(\theta) = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right| tan ( ΞΈ ) = β 1 + m 1 β m 2 β m 1 β β m 2 β β β
Summary of Steps
Calculate the slopes of the sides of the triangle:
m 1 m_1 m 1 β m 2 m_2 m 2 β m 3 m_3 m 3 β
Use the tangent formula to calculate the angles:
Angle at A: tan β‘ ( β A ) = 41 64 β β A β 32.6 4 β \tan(\angle A) = \frac{41}{64} \quad \Rightarrow \quad \angle A \approx 32.64^\circ tan ( β A ) = 64 41 β β β A β 32.6 4 β
Angle at B: tan β‘ ( β B ) = 41 11 β β B β 105.0 2 β \tan(\angle B) = \frac{41}{11} \quad \Rightarrow \quad \angle B \approx 105.02^\circ tan ( β B ) = 11 41 β β β B β 105.0 2 β
Angle at C: tan β‘ ( β C ) = 41 25 β β C β 42.3 4 β \tan(\angle C) = \frac{41}{25} \quad \Rightarrow \quad \angle C \approx 42.34^\circ tan ( β C ) = 25 41 β β β C β 42.3 4 β
Final Angles of the triangle:
β A β 32.6 4 β \angle A \approx 32.64^\circ β A β 32.6 4 β β B β 105.0 2 β \angle B \approx 105.02^\circ β B β 105.0 2 β β C β 42.3 4 β \angle C \approx 42.34^\circ β C β 42.3 4 β
