Sine - HiSet: High School Equivalency Test: Math

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Question

$\cos N = \frac{x}{3}$

Evaluate $\sin N$ in terms of .

Answer

Suppose we allow be the lengths of the opposite leg and adjacent leg and the hypotenuse, respectively, of the right triangle with an acute angle measuring . The cosine is defined to be the ratio of the length of the adjacent side to that of the hypotenuse, so

$\cos N = \frac{A}{H} = \frac{x}{3}$

We can set the lengths of the adjacent leg and the hypotenuse to and 3, respectively. By the Pythagorean Theorem, the length of the opposite leg is

$O = \sqrt{H^{2}- A^{2}} = \sqrt{x^{2}-3^{2}} = \sqrt{x^{2}-9 }$

The sine of the angle is equal to the ratio of the length of the opposite leg to that of the hypotenuse, so

$\sin N = \frac{O}{H} = \frac{\sqrt{x^{2}-9 }}{3}$ .

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Question

$\tan N = \frac{\sqrt{y}}{5}$

Evaluate $\sin N$ in terms of .

Answer

Suppose we allow be the lengths of the opposite leg and adjacent leg and the hypotenuse, respectively, of the right triangle with an acute angle measuring .

The tangent of the angle is the ratio of the length of the opposite leg to that of the adjacent leg, so

$\tan N = \frac{O}{A} =\frac{\sqrt{y}}{5}$

We can set the lengths of the opposite and adjacent legs to $\sqrt{y}$ and 5, respectively. The length of the hypotenuse can be determined using the Pythagorean Theorem:

$H = \sqrt{O^{2}+A^{2}} = \sqrt{( \sqrt{y})^{2}+5^{2}} = \sqrt{y+25}$

The sine of the angle is equal to the ratio of the length opposite leg to that of the hypotenuse, so

$\sin N = \frac{O}{H} = \frac{\sqrt{y }}{ \sqrt{y+25}} =\sqrt{ \frac{y}{y+25}}$ .

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