If $sin A = \frac{1}{3}$ $sin A = 1/3$ and 0< A < 90, then what is cos A?

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Answer 1 · 2018-06-04T13:07:04

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If $sin A = \frac{1}{3}$ $sin A=1/3$ and $0 < A < 90$ $0 < A < 90$

Then using ${sin}^{2} A + {cos}^{2} A = 1$ $sin^2 A+cos^2A=1$

${cos}^{2} A = 1 - {sin}^{2} A = 1 - \frac{1}{9} = \frac{8}{9}$ $cos^2A=1-sin^2A=1-1/9=8/9$

Then $cos A = \pm \sqrt{\frac{8}{9}} = \pm \frac{2 \sqrt{2}}{3}$ $cosA=+-sqrt(8/9)=+-(2sqrt2)/3$

But if 0< A< 90, then the cosine is positive and $cos A = \frac{2 \sqrt{2}}{3}$ $cosA=(2sqrt2)/3$

Answer 2 · 2018-06-04T13:14:19

$cos A = \frac{2 \sqrt{2}}{3}$ $cos A = (2sqrt2)/3$

Since $0 < A < 90^{o}$ $0 < A < 90^o$ we know that $∠ A$ $angle A$ is in the first quadrant.

Consider the right $△ O A B$ $triangle OAB$ where $O$ $O$ is the origin and side $A B$ $AB$ is opposite $∠ A$ $angle A$

We are told that $sin A = \frac{1}{3}$ $sin A =1/3$

$:.$ side $AB =1$ and side $OA =3$

Applying Pythagoras

$(OA)^2 = (AB)^2 + (OB)^2$

$:. 3^2 = 1^2 +OB^2$

$OB = sqrt(9-1)$ [Since $OB>0$ because $angle A$ is in the first quadrant]

$OB = sqrt8 = 2sqrt2$

Now, $cos A = (OB)/(OA)$

$cos A = (2sqrt2)/3$

If sin A = 1/3sinA=13sin A = 1/3 and 0< A < 90, then what is cos A?