find the x in figure

Geometry Level pending


The answer is 3.

Aly Ahmed by Aly Ahmed , 34, Egypt

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3 solutions

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cos ⁡ 126 ° = y 2 + ( 3 x − 4 ) 2 + ( 4 x − y ) 2 − ( 5 x − 2 ) 2 2 z ( 4 x − y ) \cos 126\degree=\dfrac{y^2+(3x-4)^2+(4x-y)^2-(5x-2)^2}{2z(4x-y)} ,

where z = y 2 + ( 3 x − 4 ) 2 z=\sqrt {y^2+(3x-4)^2} .

cos ⁡ 126 ° = − sin ⁡ 36 ° = − y z ⟹ − 2 y ( 4 x − y ) = y 2 + ( 3 x − 4 ) 2 + ( 4 x − y ) 2 − ( 5 x − 2 ) 2 ⟹ ( y + 4 x − y ) 2 + ( 3 x − 4 ) 2 − ( 5 x − 2 ) 2 = 0 ⟹ − 4 x + 12 = 0 ⟹ x = 3 \cos 126\degree= -\sin 36\degree=-\dfrac{y}{z}\implies -2y(4x-y)=y^2+(3x-4)^2+(4x-y)^2-(5x-2)^2\implies (y+4x-y)^2+(3x-4)^2-(5x-2)^2=0\implies -4x+12=0\implies x=\boxed 3 .

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By cosine rule on the triangle with the 12 6 ∘ 126^\circ angle, we have:

( 5 x − 2 ) 2 = ( 3 x − 4 ) 2 + y 2 + ( 4 x − y ) 2 − 2 ( 4 x − y ) × 3 x − 4 cos ⁡ 3 6 ∘ cos ⁡ 12 6 ∘ As cos ⁡ θ = sin ⁡ ( 9 0 ∘ − θ ) 25 x 2 − 20 x + 4 = 9 x 2 − 24 x + 16 + y 2 + 16 x 2 − 8 x y + y 2 − 2 ( 4 x − y ) × 3 x − 4 cos ⁡ 3 6 ∘ sin ⁡ ( − 3 6 ∘ ) = 9 x 2 − 24 x + 16 + y 2 + 16 x 2 − 8 x y + y 2 + 2 ( 4 x − y ) ( 3 x − 4 ) tan ⁡ 3 6 ∘ Note: tan ⁡ 3 6 ∘ = y 3 x − 4 = 9 x 2 − 24 x + 16 + y 2 + 16 x 2 − 8 x y + y 2 + 2 ( 4 x − y ) ( 3 x − 4 ) × y 3 x − 4 25 x 2 − 20 x + 4 = 9 x 2 − 24 x + 16 + y 2 + 16 x 2 − 8 x y + y 2 + 8 x y − 2 y 2 4 x = 12 ⟹ x = 3 \small \begin{aligned} (5x-2)^2 & = (3x-4)^2+y^2 + (4x-y)^2 - 2(4x-y)\times \frac {3x-4}{\cos 36^\circ} \blue{\cos 126^\circ} & \small \blue{\text{As }\cos \theta = \sin (90^\circ - \theta)} \\ 25x^2 - 20x + 4 & = 9x^2 - 24x + 16 + y^2 + 16x^2 - 8xy + y^2 - 2(4x-y)\times \frac {3x-4}{\cos 36^\circ} \blue{\sin (-36^\circ) } \\ & = 9x^2 - 24x + 16 + y^2 + 16x^2 - 8xy + y^2 + 2(4x-y)(3x-4)\blue{\tan 36^\circ} & \small \blue{ \text{Note: }\tan 36^\circ = \frac y{3x-4}} \\ & = 9x^2 - 24x + 16 + y^2 + 16x^2 - 8xy + y^2 + 2(4x-y)\cancel{(3x-4)}\times \blue{\frac y{\cancel{3x-4}}} \\ \cancel{25x^2} - 20x + 4 & = \cancel{9x^2} - 24x + 16 + \cancel{y^2} + \cancel{16x^2} - \cancel{8xy} + \cancel{y^2} + \cancel{8xy} - \cancel{2y^2} \\ 4x & = 12 \\ \implies x & = \boxed 3 \end{aligned}

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