If `sin A =3/4` , calculate cos A and tan A.

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#### Solution

Let ΔABC be a right-angled triangle, right-angled at point B.

Given that,

sin A = 3/4

`(BC)/(AC) = 3/4`

Let BC be 3*k*. Therefore, AC will be 4*k*, where *k* is a positive integer.

Applying Pythagoras theorem in ΔABC, we obtain

AC^{2} = AB^{2} + BC^{2}

(4*k*)^{2} = AB^{2} + (3*k*)^{2}

16*k* ^{2} − 9*k* ^{2} = AB^{2}

7*k* ^{2} = AB^{2}

AB = sqrt7k

`cos A = ("Side adjacent to"angleA)/"Hypotenuse"`

=`(AB)/(AC) = sqrt(7k)/(4k) = sqrt7/4`

`tan A = ("Side adjacent to"angleA)/("Side adjacent to"angleA)`

`= (BC)/(AB) = (3k)/(sqrt7k) = 3/sqrt7`

Concept: Trigonometric Ratios

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