A Very Nice Geometry Problem | You should be able to solve this | 2 Methods

A Very Nice Geometry Problem | You should be able to solve this | 2 Methods
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[@indirachandran2785]

How can we construct a line which divides an angle 3theta into 2 angles where as theta is to be determined. Any line perpendicular we can construct which we are sure about the angle. We can draw a line DE such that DA=DEthat make sense I think.

[@user-ii5vr3um9r]

Alternatively, find M on AB so that angle MDA=theta, so AM=MD=DB, suppose AM=a. Find N on DC so that DN=BD, so BM=CN and angle BMN=90, suppose BM=b. BM:BN=AB:BC, so b=a*sqrt(2), so theta=45/2=22.5

[@richardleveson6467]

Clever problem and clever solutions!

[@misterenter-iz7rz]

As usual, sine rule works for the type of problem, eventually we get sin 5 t=sin 3t, so 8t=180, t=180/8=22.5.

[@yakupbuyankara5903]

22.5 degrees

[@ai2657]

sir SIN(3θ) =SIN(5θ) WE FIND ANS IS 22.5 OR 67.5 BYT 67.5 REJECTED ANS IS 22.5

[@giuseppemalaguti435]

Usando sempre il teorema dei seni risulta sin3θ/cosθ=tg2θ.. sviluppando i calcoli risulta tgθ=√(3-√8)..θ=22,5

[@guruputramali7033]

Sir different method

[@kannanganapathi9403]

Can't understand last step. Suddenly 5 + 3 = 180 how?

[@guruputramali7033]

Difficult 😢

[@misterenter-iz7rz]

36

[@holyshit922]

cos(2theta) = x /(y+x) , ABD right triangle From law of sines in triangle ABD y/sin(theta) = x/sin(180°-3theta) y = sin(theta)/sin(3theta)*x sin(3theta)=sin(theta)cos(2theta)+cos(theta)sin(2theta) sin(3theta)=sin(theta)(2cos^2(theta) - 1)+2sin(theta)cos^2(theta) sin(3theta)=sin(theta)(2cos^2(theta) - 1+2cos^2(theta)) sin(3theta)=sin(theta)(4cos^2(theta) - 1) y = x/(4cos^2(theta) - 1) y = x/(2*(2cos^2(theta) - 1)+1) y = x/(2cos(2theta)+1) cos(2theta) = x/(x/(2cos(2theta)+1)+x) c = x/((1/(2c+1)+1)x) c = 1/((2c+2)/(2c+1)) c = (2c+1)/(2c+2) c(2c+2) = (2c+1) 2c^2+2c-2c-1=0 c^2 = 1/2 c = 1/sqrt(2) theta < 45° 2theta = 45° theta = 22°30'

[@mauriciosahady]

First method: don't need line DE