Question

if sin A = 3/4 then calculate cos A and tan A
​

Answer 1

Step-by-step explanation:

hope it's help uuu....

Answer 2

Solution:

Let us say, ABC is a right-angled triangle, right-angled at B.

➢ Sin A = 3/4

As we know,

Sin A = Opposite Side/Hypotenuse Side = 3/4

$\bullet{\sf{\underline{ Diagram:}}}$

$\setlength{\unitlength}{1.6mm}\begin{picture}(30,20)\put(-3,-3){\line(1,1){12}}\put(-3,-3){\line(1,0){20}}\put(17,-3){\line(-2,3){8}}\end{picture}$

Step By step:

Now,

let BC be 3k and AC will be 4k.

[where k is the positive real number.]

According to the Pythagoras theorem;

we know,

(Hypotenuse² = Perpendicular²+ Base²)

➢ (h)² = (p)² + (b)²

➢ AC² = AB² + BC²

Substitute the value of AC and BC in the above expression to get;

➢(4k)² = (AB)² + (3k)²

➢16k² – 9k² = AB²

➢AB² = 7k²

Hence,

AB = √7 k

We need to Find Values:

cos A = Adjacent Side/Hypotenuse side = AB/AC

cos A = √7 k/4k = √7/4

And,

tan A = Opposite side/Adjacent side = BC/AB

tan A = 3k/√7 k = 3/√7

Thanks: