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

$\frac{19}{\sqrt{50}}$

tan²A + 2tanA - 63 = 0

tan²A +9tanA - 7tanA - 63 = 0

on solving ,

tanA = \(\frac{7}{1}\)

By using pythagoras theorem ,

P² + B² = H²

7² + 1² = H²

H = √50

Now , ( 2sinA + 5cosA)

= 2 ( \(\frac{7}{√50}\)  ) + 5 ( \(\frac{1}{√50}\)  )

=  \(\frac{19}{√50}\)