But it's not a right triangle!

Geometry Level 2

Express the value of tan ⁡ θ \tan\theta in terms if x x and y y .

y x \frac yx y x + y \frac y{x+y} x x + y \frac x{x+y} y 2 x + y \frac{y}{2x+y}

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Ab Awan by AB Awan , 29, Pakistan

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2 solutions

Alan Guo
Nov 27, 2015

tan ⁡ π / 2 + θ = tan ⁡ π / 2 + tan ⁡ θ 1 − tan ⁡ π / 2 tan ⁡ θ \tan{\pi/2 + \theta} = \dfrac{\tan{\pi/2} +\tan{\theta}}{1-\tan{\pi/2}\tan{\theta}} x + y x = 1 + tan ⁡ θ 1 − tan ⁡ θ \dfrac{x+y}{x}=\dfrac{1+\tan{\theta}}{1-\tan{\theta}} x + y − ( x + y ) tan ⁡ θ = x + x tan ⁡ θ x+y -(x+y)\tan{\theta} = x + x\tan{\theta} y = ( 2 x + y ) tan ⁡ θ y = (2x+y)\tan{\theta} tan ⁡ θ = y 2 x + y \tan{\theta}=\dfrac{y}{2x+y}

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i think you meant π 4 \dfrac{\pi}{4} , anyways same solution.

Aareyan Manzoor - 5 years, 6 months ago

By the Law of Sines:

y sin ⁡ θ = x 2 sin ⁡ ( 4 5 o − θ ) \displaystyle{\frac{y}{\sin \theta}=\frac{x\sqrt{2}}{\sin (45^o-\theta)}}

sin ⁡ ( 4 5 o − θ ) = sin ⁡ 4 5 o cos ⁡ θ − sin ⁡ θ cos ⁡ 4 5 o \displaystyle{\sin (45^o-\theta)= \sin 45^o \cos \theta - \sin \theta \cos 45^o}

sin ⁡ θ = y ( sin ⁡ 4 5 o cos ⁡ θ − sin ⁡ θ cos ⁡ 4 5 o ) x 2 \displaystyle{\sin \theta = \frac{y(\sin 45^o \cos \theta - \sin \theta \cos 45^o)}{x \sqrt{2}}}

sin ⁡ θ = y ( 1 2 cos ⁡ θ − sin ⁡ θ 1 2 ) x 2 \displaystyle{\sin \theta = \frac{y(\frac{1}{\sqrt{2}} \cos \theta - \sin \theta \frac{1}{\sqrt{2}})}{x \sqrt{2}}}

cos ⁡ θ y 2 x − sin ⁡ θ y 2 x = sin ⁡ θ \displaystyle{\frac{\cos \theta y}{2x}-\frac{\sin \theta y}{2x} = \sin \theta}

Dividing the equation by cos ⁡ θ \cos \theta :

y 2 x − tan ⁡ θ y 2 x = tan ⁡ θ \displaystyle{\frac{y}{2x}-\tan \theta \frac{y}{2x} = \tan \theta}

y 2 x = tan ⁡ θ + tan ⁡ θ y 2 x \displaystyle{\frac{y}{2x} = \tan \theta+\tan \theta \frac{y}{2x}}

y 2 x = tan ⁡ θ ( 1 + y 2 x ) \displaystyle{\frac{y}{2x} = \tan \theta (1+\frac{y}{2x})}

y 2 x 2 x 2 x + y = tan ⁡ θ \displaystyle{\frac{y}{2x} \frac{2x}{2x+y}= \tan \theta}

tan ⁡ θ = y 2 x + y \displaystyle{\tan \theta= \frac{y}{2x+y} }

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