If tan2A + 2tanA - 63 = 0 Given that 0 < A < $\frac{\pi}{2}$ what is the value of ( 2sinA + 5cosA)?
Answer & explanation
Correct answer: option 3
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 ( \(\frac{1}{√50}\) )
= \(\frac{19}{√50}\)