Mathematics Questions and Answers

PR² = PQ² + QR² - 2(QR)(PQ) COS 120^o

PR² = 1² + 2² - 2(1)(2) x - cos 60^o

= 5 - 2(1)(2) x -\(\frac{1}{2}\)

= 5 + 2 = 7

PR = \(\sqrt{7}\)cm

\(\cot \theta = \frac{x}{y}\)

\(\implies \tan \theta = \frac{y}{x}\)

\(opp = y; adj = x\)

Using Pythagoras theorem, \(Hyp^{2} = Opp^{2} + Adj^{2}\)

\(Hyp^{2} = y^{2} + x^{2}\)

\(Hyp = \sqrt{y^{2} + x^{2}}\)

\(\sin \theta = \frac{y}{\sqrt{y^{2} + x^{2}}}\)

\(\therefore \csc \theta = \frac{\sqrt{y^{2} + x^{2}}}{y}\)

= \(\frac{1}{y}(\sqrt{y^{2} + x^{2}})\)

\(\frac{4a^2 - 49b^2}{2a^2 - 5ab - 7b^2}\) = \(\frac{(2a)^2 - (7b)^2}{(a + b)(2a - 7b)}\)

= \(\frac{(2a + 7b)(2a - 7b)}{(a + b)(2a - 7b)}\)

= \(\frac{2a + 7b}{a + b}\)

\(\frac{1}{x^2 + 5x + 6}\) + \(\frac{1}{x^2 + 3x + 2}\) = \(\frac{1}{(x + 1)(x + 2)}\) + \(\frac{1}{(x + 1)(x + 1)}\)

\(\frac{(x + 1)+ (x + 3)}{(x + 1)(x + 2)(x + 3)}\) = \(\frac{x + 1 + x + 3}{(x + 1)(x + 2)(x + 3)}\)

\(\frac{2x + 4}{(x + 1)(x + 2)(x + 3)}\) = \(\frac{2(x + 2)}{(x + 1)(x + 2)(x + 3)}\)

= \(\frac{2}{(x + 1)(x + 3)}\)

px² + qx + b = 0

Using almighty formula

\(\frac{-b \pm \sqrt{ b^2 - 4ac}}{2a}\).........(i)

Where a = p, b = q and c = b

substitute for this value in equation (i)

= \(\frac{-q \pm \sqrt{ q^2 - 4bp}}{2p}\)