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Right Triangles and Trigonometry

SAT Math· difficulty 3/5

In right triangle ABCABC, angle C=90°C=90°. If sin⁡A=25\sin A = \frac{2}{\sqrt{5}}, what is cos⁡B\cos B?

  • A

    52\frac{\sqrt{5}}{2}

  • B

    15\frac{1}{\sqrt{5}}

  • C

    12\frac{1}{2}

  • D

    25\frac{2}{\sqrt{5}}

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Explanation

Since A+B=90°A+B=90°, cos⁡B=sin⁡A=25\cos B=\sin A=\frac{2}{\sqrt{5}}.

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