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hey buddy,
the answer of this question will be a.
c=a sin@ + b Cos@
c/√(a^2+b^2)= (a sin@ /√(a^2+b^2) + (b cos@/√(a^2+ b^2)
If u take a right angled triangle witrh sides a, b then its hypotenues will be √(a^2+b^2)
in this triangle let the other angle be $
therefore its in the form of:
c/√(a^2+b^2) = cos $ sin @ + sin$ cos @
since, sin(@+$) = cos $ sin @ + sin$ cos @
c= √(a^2+ b^2) sin($+@)
and $ + @ = 90
therefore
a sin@ + b cos $ will be
√(a^2+ b^2)
hope it will help
the answer of this question will be a.
c=a sin@ + b Cos@
c/√(a^2+b^2)= (a sin@ /√(a^2+b^2) + (b cos@/√(a^2+ b^2)
If u take a right angled triangle witrh sides a, b then its hypotenues will be √(a^2+b^2)
in this triangle let the other angle be $
therefore its in the form of:
c/√(a^2+b^2) = cos $ sin @ + sin$ cos @
since, sin(@+$) = cos $ sin @ + sin$ cos @
c= √(a^2+ b^2) sin($+@)
and $ + @ = 90
therefore
a sin@ + b cos $ will be
√(a^2+ b^2)
hope it will help
joshihimanshu46:
do u know the range of value of sin and cos
if theta will be 90 then sin will be 1 but cosine will become zero
ok
i thought the value of theta can be different
OK if you get something plz tell me
i am going to edit the ans
ya a is the right ans
somebody have reported my ans
i can't edit it
hey buddy its done
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