यदि `sin A= (3)/(4) ` तो cos A और tan A का मान परिकलित कीजिए |

Explanation

` sin A = (3)/(4)`,में
`Delta ABC `
माना BC = 3k तथा AC = 4k
` therefore AB^(2) + BC^(2) = AC^(2) rArr AB^(2) = AC^(2) - BC^(2) `
`rArr AC^(2) (4k)^(2) - (3k)^(2)`
` = 16 k^(2) - 9k^(2) = 7k^(2) rArr AB = sqrt(7) k`
अब `cos A = ("आधार ")/(" कर्ण ") (sqrt(7)k)/(4k) = (sqrt(7))/(4)`
तथा `tan A = ("लम्ब ")/("आधार ") = (3k)/(sqrt(7)k) = (3)/(sqrt(7))`