Question

sinA=4/5 find cosA and tanA​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

sinA=4/5.

Since it is a ratio, lets assume it as 4x and 5x.

Lets assume a triangle (Given in file attached below)

Lets sinA=Opposite/hypotenuse

SinA=BC/AC

But SinA=4/5

∴BC/AC=4/5

thus, lets assume BC=4x, AC=5x.

Using pythagoras theorem,

AC²=AB²+BC²

AB²=AC²-BC²

AB²=(5x)²-(4x)²

AB=√(9x²)

AB=3x.

Thus, now its easy to find tan and cos...

CosA=adjacent/opposite

cosA=AB/AC

cosA=3x/5x..................x and x gets cancelled

cosA=3/5

Similarly,

tanA=opposite/adjacent

tanA=BC/AB

tanA=4x/3x

tanA=4/3.

HOPE THIS HELPS :D