Given that sinA=a/b, then find the value of cosA.
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Answered by
1
Answer:
root of b^2 - a^2\b^2
Step-by-step explanation:
cosA = root of 1 - sinA^2
= root of b^2 - a^2
---------------------
a^2
Answered by
1
SinA=a/b
Assuming triangle ABC where AC is the hypotenuse
In triangle ABC
AC2=AB2+BC2
Sin=opposite/hypotenuse
Assuming Angle A as theta
SinA= BC/AC = a/b
(b)2=AB2+(a)2
b2-a2=AB2
AB=root b2-a2
CosA=AB/AC
CosA=root b2-a2/b
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Assuming triangle ABC where AC is the hypotenuse
In triangle ABC
AC2=AB2+BC2
Sin=opposite/hypotenuse
Assuming Angle A as theta
SinA= BC/AC = a/b
(b)2=AB2+(a)2
b2-a2=AB2
AB=root b2-a2
CosA=AB/AC
CosA=root b2-a2/b
_______________
Mark as brainliest and thank me........................
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