This note describes about Simple Problems in Mathematics.
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1. The Learning Hall Mathematics Class sep 30, 2018 Find the co-ordinates of the orthocentre of the triangle whose vertices are (-1,3), (2, -1), and (0,0) A (-1,3) c (0,0) D In the above figure A (-1,3), B (2, -1), and C (0,0) are the vertices of A ABC. P is the concurrent point of altitude, so it is orthocentre. Now, slope of BC = ml = (Y2-Y1)/ =( 0-(-1))/(0-2) = 1/-2 Since, AD is perp. to BC, Slope of AD = m-z= -l/nu= 2 Hence, equation of AD Is Y2-Y1 = or, y-3 = 2 (x- or, y-3 = 2 0
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Similarly, slope of AC= ml = (Y2-Y1)/ - Since, BE is perp. to AC, Slope of BE = -l/ml= 1/3 Hence, equation of BE Y2-Y1 = Is or, y-(-1) = 1/3 (x -2) or, Y+l = 1/3 (x-2) or, x-3y = 5 (ii) On solving eq. (i) & (ii), we get x = - 4, and y = -3 Co-ordinates of Ortho Centre = 2. Prove that p (-4,-3) (Cos4x + Cos 3x + Cos 2x)/(Sin 4x +Sin 3x+ Sin 2x) = A. Now, Cot 3x LHS = (Cos4x + Cos 3x + Cos 2x)/(Sin 4x +Sin 3x+ Sin 2x) = 2Cos ((4X+2x)/2).Cos ((4x-2x)/2)+ Cos 3x 2Sin ((4X+2x)/2).Cos ((4x-2x)/2)+ Sin 3x = 2 Cos 3x. Cos x + Cos 3x 2 Sin 3x. Cos x + Sin 3x = Cos 3x.(2Cos x + 1)
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3. A. Sin 3x. (2Cos x + 1) = Cot = RI-IS If x + iy = J (a + ib)/(p+iq), then prove that (x2+Y) - 22 (a2+ b2)/ (p2+ q2) Now, given that x + iy = VI (a + ib)/(p+iq) where I = imaginary number = (-1) Hence, x- iy = J (a - ib)/(p-iq) (ii) By (i) x (ii), we get (x + iy).( x— iy) = (a + ib)/(p+iq). VI (a - ib)/(p-iq) = VI (a + ib).(a - ib)/ v/(p+iq).(p-iq) Or + Y2 = (a2 + b2)/ J(p2+q2) Or (x2 + Y ) 2 2 = (a2 + b2)/ (p2+q2)
Discussion
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