Solution 4.2:1a
From Förberedande kurs i matematik 1
(Difference between revisions)
			  			                                                      
		          
			m   | 
			|||
| (3 intermediate revisions not shown.) | |||
| Line 1: | Line 1: | ||
| - | {{  | + | The definition of the tangent states that  | 
| - | <  | + | |
| - | {{  | + | {| width="100%"  | 
| - | + | | width="50%" align="center"|<math>\tan u=\frac{\text{opposite}}{\text{adjacent}}</math>  | |
| + | | width="50%" align="center"|[[Image:4_2_1_a.gif]]  | ||
| + | |}  | ||
| + | |||
| + | In our case, this means that  | ||
| + | |||
| + | {{Displayed math||<math>\tan 27^{\circ} = \frac{x}{13}</math>}}  | ||
| + | |||
| + | which gives <math>x = 13\cdot \tan 27^{\circ}\,</math>.  | ||
| + | |||
| + | |||
| + | Note: Using a calculator, we can work out what ''x'' should be,  | ||
| + | |||
| + | {{Displayed math||<math>x = 13\cdot\tan 27^{\circ} \approx 6\textrm{.}62\,\textrm{.}</math>}}  | ||
Current revision
The definition of the tangent states that
| \displaystyle \tan u=\frac{\text{opposite}}{\text{adjacent}} |  
 | 
In our case, this means that
| \displaystyle \tan 27^{\circ} = \frac{x}{13} | 
which gives \displaystyle x = 13\cdot \tan 27^{\circ}\,.
Note: Using a calculator, we can work out what x should be,
| \displaystyle x = 13\cdot\tan 27^{\circ} \approx 6\textrm{.}62\,\textrm{.} | 

