Solution 4.2:1d
From Förberedande kurs i matematik 1
(Difference between revisions)
			  			                                                      
		          
			m  (Lösning 4.2:1d moved to Solution 4.2:1d: Robot: moved page)  | 
				|||
| Line 1: | Line 1: | ||
| - | {{NAVCONTENT_START}}  | ||
| - | <center> [[Image:4_2_1d.gif]] </center>  | ||
| - | {{NAVCONTENT_STOP}}  | ||
[[Image:4_2_1_d.gif|center]]  | [[Image:4_2_1_d.gif|center]]  | ||
| + | |||
| + | The side marked   | ||
| + | <math>x</math>  | ||
| + | is the hypotenuse in the right-angled triangle and the side of length   | ||
| + | <math>\text{16}</math>  | ||
| + | is the adjacent to the angle of  | ||
| + | <math>\text{2}0^{\circ }</math>.  | ||
| + | |||
| + | |||
| + | By writing the quotient for   | ||
| + | <math>\text{cos20}^{\circ }</math>, we obtain the relation  | ||
| + | |||
| + | |||
| + | <math>\text{cos20}^{\circ }=\frac{16}{x}</math>  | ||
| + | |||
| + | |||
| + | and this gives  | ||
| + | |||
| + | |||
| + | <math>x=\frac{16}{\text{cos20}^{\circ }}\quad \left( \approx 17.0 \right).</math>  | ||
Revision as of 11:08, 28 September 2008
The side marked \displaystyle x is the hypotenuse in the right-angled triangle and the side of length \displaystyle \text{16} is the adjacent to the angle of \displaystyle \text{2}0^{\circ }.
By writing the quotient for 
\displaystyle \text{cos20}^{\circ }, we obtain the relation
\displaystyle \text{cos20}^{\circ }=\frac{16}{x}
and this gives
\displaystyle x=\frac{16}{\text{cos20}^{\circ }}\quad \left( \approx 17.0 \right).

