Question

hey mate
plz solve this
urgent
sinx=0.6 then sin90-x​

Answer 1

Given since=0.6

sin(90-x)=cosx

we know that sin^2(x)+cos^2(x)=1

(0.6)^2 + cos^2 (x)= 1

0.36+ cos^2 (x)=1

cos^2(x) = 0.64

cosx= 0.8

cos x = sin (90-x) = 0.8

Answer 2

ANSWER:

$\sin(x) = 0.6 - - - (1) \\we \: know \\ = > \sin(90 - x) = cos(x) - - - (2) \\ from \: (1) \\ {sin}^{2}x = {(0.6)}^{2} \\ = > {sin}^{2} x = 0.36 \\ we \: know \\ {sin}^{2}x = 1 - {cos}^{2} x \\ = > 0.36 - 1 = - {cos}^{2} x \\ = > 1 - 0.36 = {cos}^{2} x \\ = > {cos}^{2} x = 0.64 \\ = > cos \: x \: = 0.8 \\ = > sin(90 - x) = 0.8$

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