Question

If sin A = ¾, calculate cos A and tan A.​

Answer 1

Given, sinA=

4

3

⇒

DC

BC

=

4

3

⇒BC=3k and AC=4k

where k is the constant of proportionality.

By Pythagoras theorem, we have

AB

2

=AC

2

−BC

2

=(4k)

2

−(3k)

2

=7k

2

⇒AB=

7

k

So, cosA=

AC

AB

=

4k

7

k

=

4

7

And tanA=

AB

BC

=

7

k

3k

=

7

3

p

Answer 2

Answer:

If sin A = ¾, calculate cos A and tan A.

Step-by-step explanation:

SinA = 3/4

Let the ratio be in the X.

perpendicular = 3x

hypotenuse = 4x

$base \: = \sqrt{(4x)^{2} - (3x)^{2}} = \sqrt{16 {x}^{2} - 9 {x}^{2} } = \sqrt{7 {x}^{2} } = \sqrt{7} x \\ cos \: a = \frac{b}{h} = \frac{ \sqrt{7}x }{4x} = \frac{ \sqrt{7} }{4} \\ tan \: a = \frac{p}{b} = \frac{3x}{ \sqrt{7}x } = \frac{3}{ \sqrt{7} }$

cos A = √7/4

tanA = 3/√7