Solution 4.3:5

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An often-used technique to calculate
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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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<math>\text{cos }v</math>
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and
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<math>\text{tan }v</math>, given the sine value of an acute angle, is to draw the angle
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<math>v</math>
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in a right-angled triangle which has two sides arranged so that
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<math>\text{sin }v={5}/{7}\;</math>.
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[[Image:4_3_5_1.gif|center]]
[[Image:4_3_5_1.gif|center]]
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Using Pythagoras' theorem, we can determine the length of the third side in the triangle.
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Using the Pythagorean theorem, we can determine the length of the third side in the triangle.
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{| align="center"
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[[Image:4_3_5_2.gif|center]]
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||[[Image:4_3_5_2.gif]]
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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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|}
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<math>x^{2}+5^{2}=7^{2}</math>
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which gives that
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<math>x=\sqrt{7^{2}-5^{2}}=\sqrt{24}=2\sqrt{6}</math>
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Then, using the definition of cosine and tangent,
Then, using the definition of cosine and tangent,
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{{Displayed math||<math>\begin{align}
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<math>\begin{align}
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\cos v &= \frac{x}{7} = \frac{2\sqrt{6}}{7}\,,\\[5pt]
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& \cos v=\frac{x}{7}=\frac{2\sqrt{6}}{7}, \\
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\tan v &= \frac{5}{x} = \frac{5}{2\sqrt{6}}\,\textrm{.}
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& \tan v=\frac{5}{x}=\frac{5}{2\sqrt{6}} \\
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\end{align}</math>}}
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\end{align}</math>
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NOTE: Note that 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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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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