If 4cosA + 3secA = 4√3, 0° < 'A' < 90°, then the value of 'A' is:
Show answer & explanation
Correct answer: C
Explanation
Given:
4cosA + 3secA = 4√3
0° < A < 90°
By formula:
secA = 1 / cosA
4cosA + 3/cosA = 4√3
Let cosA = x
4x + 3/x = 4√3
Multiplying by x
4x² + 3 = 4√3 x
4x² − 4√3 x + 3 = 0
a = 4, b = −4√3, c = 3
x = (4√3 ± √((4√3)² − 4×4×3)) / (2×4)
= (4√3 ± √(48 − 48)) / 8
= 4√3 / 8
= √3 / 2
Therefore
cosA = √3 / 2
A = 30°