Solution 4.2:1e

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[[Image:4_2_1_e.gif|center]]
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In the triangle, we seek the hypotenuse
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<math>x</math>, knowing the angle 35o and that the adjacent has length 11.
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The definition of sine gives
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<math>\sin 35^{\circ }=\frac{11}{x}</math>
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and thus
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<math>x=\frac{11}{\sin 35^{\circ }}\quad \left( \approx 19.2 \right)</math>

Revision as of 11:12, 28 September 2008

In the triangle, we seek the hypotenuse \displaystyle x, knowing the angle 35o and that the adjacent has length 11.


The definition of sine gives


\displaystyle \sin 35^{\circ }=\frac{11}{x}


and thus


\displaystyle x=\frac{11}{\sin 35^{\circ }}\quad \left( \approx 19.2 \right)

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