Question

If sinA=3/5 then find the value of cosA
(a) 1. (b) 4/5. (c)3/4. (d) 5/3​

Answer 1

Answer:

B)4/5

Step-by-step explanation:

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Answer 2

given, sinA=3/5=BC/AC

Let, BC=k and AC=k

AB2=AC2-BC2

=5k2-3k2

=25k2-9k2

=16k2

AB=

$\sqrt{16k {}^{2} } = 4k$

CosA=AB/AC

=4k/5k=4/5

so correct option is (b)