Solution 4.3:5

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An often-used technique to calculate <math>\cos v</math> and <math>\tan v</math>, given the sine value of an acute angle, is to draw the angle <math>v</math> in a right-angled triangle which has two sides arranged so that <math>\sin v = 5/7\,</math>.
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Using the Pythagorean theorem, we can determine the length of the third side in the triangle.
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||<math>\begin{align}&x^2 + 5^2 = 7^2\\[5pt] &\text{which gives that}\\[5pt] &x = \sqrt{7^2-5^2} = \sqrt{24} = 2\sqrt{6}\end{align}</math>
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Then, using the definition of cosine and tangent,
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{{Displayed math||<math>\begin{align}
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\cos v &= \frac{x}{7} = \frac{2\sqrt{6}}{7}\,,\\[5pt]
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\tan v &= \frac{5}{x} = \frac{5}{2\sqrt{6}}\,\textrm{.}
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\end{align}</math>}}
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Note: The right-angled triangle that we use is just a tool and has nothing to do with the triangle that is referred to in the question.

Current revision

An often-used technique to calculate \displaystyle \cos v and \displaystyle \tan v, given the sine value of an acute angle, is to draw the angle \displaystyle v in a right-angled triangle which has two sides arranged so that \displaystyle \sin v = 5/7\,.

Using the Pythagorean theorem, we can determine the length of the third side in the triangle.

Image:4_3_5_2.gif \displaystyle \begin{align}&x^2 + 5^2 = 7^2\\[5pt] &\text{which gives that}\\[5pt] &x = \sqrt{7^2-5^2} = \sqrt{24} = 2\sqrt{6}\end{align}

Then, using the definition of cosine and tangent,

\displaystyle \begin{align}

\cos v &= \frac{x}{7} = \frac{2\sqrt{6}}{7}\,,\\[5pt] \tan v &= \frac{5}{x} = \frac{5}{2\sqrt{6}}\,\textrm{.} \end{align}


Note: The right-angled triangle that we use is just a tool and has nothing to do with the triangle that is referred to in the question.

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