Trigonometry! 78

Geometry Level 4

Equation of line B C BC is 2 x + y − 10 = 0 2x+y-10=0 . Point A A is such that ∠ A B C = ∠ A C B = 3 0 ∘ \angle ABC = \angle ACB=30^{\circ} . Then enter the sum of the slopes of lines A B AB and A C AC .

This problem is part of the set Trigonometry .


The answer is 16.

Omkar Kulkarni by Omkar Kulkarni , 22, India

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1 solution

With X-axis, BC makes an angle of ArcTan(-2)
AB makes an angle of ArcTan(-2)+30...............AC makes an angle of ArcTan(-2)-30.
S o t h e s l o p e o f A B = − 2 + 1 3 1 − 2 3 , S o t h e s l o p e o f A C = − 2 − 1 3 1 + 2 3 . ∴ S u m o f s l o p e s o f A B a n d A C = − 2 + 1 3 1 − 2 3 + − 2 − 1 3 1 + 2 3 = − 2 + 1 3 + 4 3 − 2 3 − 2 − 1 3 − 4 3 − 2 3 1 − 4 3 = 16 So\ the\ slope\ of\ AB =\dfrac{-2+\frac 1 {\sqrt3} } {1-\frac 2 {\sqrt3} },\\ So\ the\ slope\ of\ AC =\dfrac{-2-\frac 1 {\sqrt3} } {1+\frac 2 {\sqrt3} }.\\ \therefore\ Sum\ of\ slopes\ of\ AB\ and\ AC=\dfrac{-2+\frac 1 {\sqrt3} } {1-\frac 2 {\sqrt3} }+\dfrac{-2-\frac 1 {\sqrt3} } {1+\frac 2 {\sqrt3} }\\ =\dfrac{-2+\frac 1 {\sqrt3}+\frac 4 {\sqrt3}-\frac 2 3-2-\frac 1 {\sqrt3}-\frac 4 {\sqrt3} -\frac 2 3 } {1-\frac 4 3}=\Large\ \ \color{#D61F06}{16}

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