If tan²A + 2tanA - 63 = 0, given that 0 < A < $$\frac{\pi }{2}$$ what is the value of (2sinA + 5cosA)?

Solution (By Examveda Team)

$$\eqalign{ & {\text{Given,}} \cr & {\tan ^2}A + 2\tan A - 63 = 0 \cr & {\bf{Formula \,used:}} \cr & {\text{Pythagoras Theorem,}} \cr & h = \sqrt {{p^2} + {b^2}} \cr & {\text{Where, h = Hypotenuse, p = Perpendicular and b = base}} \cr & {\bf{Calculation:}} \cr & {\tan ^2}A + 2\tan A - 63 = 0 \cr & {\text{Let, }}\tan A = x \cr & \Rightarrow {x^2} + 2x - 63 = 0 \cr & \Rightarrow {x^2} + 9x - 7x - 63 = 0 \cr & \Rightarrow x\left( {x + 9} \right) - 7\left( {x + 9} \right) = 0 \cr & \Rightarrow \left( {x + 9} \right)\left( {x - 7} \right) \cr & \Rightarrow x + 9 = 0 \Rightarrow x = - 9\left[ {'' - ''{\text{ will be neglected because }}0 < A < \frac{\pi }{2}} \right] \cr & \Rightarrow x - 7 = 0 \Rightarrow x = 7 \cr & {\text{So,}}\tan A = \frac{7}{1} = \frac{p}{b} \cr & {\text{Using Pythagoras Theorem,}} \cr & h = \sqrt {{p^2} + {b^2}} \cr & h = \sqrt {{7^2} + {1^2}} \cr & h = \sqrt {50} \cr & {\text{So}},\,2\sin A + 5\cos A \cr & = \left[ {2 \times \frac{7}{{\sqrt {50} }}} \right] + \left[ {5 \times \frac{1}{{\sqrt {50} }}} \right] \cr & = \frac{{14}}{{\sqrt {50} }} + \frac{5}{{\sqrt {50} }} \cr & = \frac{{14 + 5}}{{\sqrt {50} }} \cr & = \frac{{19}}{{\sqrt {50} }} \cr & \therefore {\text{The value of}}\left( {2\sin A + 5\cos A} \right){\text{is }}\frac{{19}}{{\sqrt {50} }}. \cr} $$