Solution 4.2:2b
From Förberedande kurs i matematik 1
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| - | <center> [[Image:4_2_2b.gif]] </center>  | ||
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[[Image:4_2_2_b.gif|center]]  | [[Image:4_2_2_b.gif|center]]  | ||
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| + | In this right-angled triangle, the opposite and the hypotenuse are given. This means that we can directly set up a relation for the sine of the angle  | ||
| + | <math>v</math>,  | ||
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| + | <math>\text{sin }v\text{ }=\text{ }{70}/{110}\;</math>.  		  | ||
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| + | The right-hand side in this equation can be simplified, so that we get  | ||
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| + | <math>\text{sin }v\text{ }=\text{ }{7}/{11}\;</math>.  | ||
Revision as of 11:26, 28 September 2008
In this right-angled triangle, the opposite and the hypotenuse are given. This means that we can directly set up a relation for the sine of the angle \displaystyle v,
\displaystyle \text{sin }v\text{ }=\text{ }{70}/{110}\;.  		
The right-hand side in this equation can be simplified, so that we get
\displaystyle \text{sin }v\text{ }=\text{ }{7}/{11}\;.

