if A is an acute angle of ABC right angle at B then the value of sin squareA+cos squareA
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ABC is a right angled triangle and right angle at B
sinA=
AC
BC
, cosA=
AC
AB
⟹sinA+cosA=
AC
BC
+
AC
AB
=
AC
BC+AB
[We know that sum of two sides of a triangle is greater than the third side]
⟹sinA+cosA>
AC
AC
=1
Hence, sinA+cosA>1
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