Solution 4.3:6b

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
m (Lösning 4.3:6b moved to Solution 4.3:6b: Robot: moved page)
Current revision (13:00, 10 October 2008) (edit) (undo)
m
 
(One intermediate revision not shown.)
Line 1: Line 1:
-
{{NAVCONTENT_START}}
+
We draw an angle <math>v</math> in the unit circle, and the fact that <math>\sin v = 3/10</math> means that its ''y''-coordinate equals <math>3/10</math>.
-
<center> [[Image:4_3_6b-1(2).gif]] </center>
+
-
{{NAVCONTENT_STOP}}
+
-
{{NAVCONTENT_START}}
+
-
<center> [[Image:4_3_6b-2(2).gif]] </center>
+
-
{{NAVCONTENT_STOP}}
+
[[Image:4_3_6_b1.gif|center]]
[[Image:4_3_6_b1.gif|center]]
 +
 +
With the information that is given, we can define a right-angled triangle in the second quadrant which has a hypotenuse of 1 and a vertical side of length 3/10.
[[Image:4_3_6_b2.gif|center]]
[[Image:4_3_6_b2.gif|center]]
 +
 +
We can determine the triangle's remaining side by using the Pythagorean theorem,
 +
 +
{{Displayed math||<math>a^2 + \Bigl(\frac{3}{10}\Bigr)^2 = 1^2</math>}}
 +
 +
which gives that
 +
 +
{{Displayed math||<math>a = \sqrt{1-\Bigl(\frac{3}{10}\Bigr)^2} = \sqrt{1-\frac{9}{100}} = \sqrt{\frac{91}{100}} = \frac{\sqrt{91}}{10}\,\textrm{.}</math>}}
 +
 +
This means that the angle's ''x''-coordinate is <math>-a</math>, i.e. we have
 +
 +
{{Displayed math||<math>\cos v=-\frac{\sqrt{91}}{10}</math>}}
 +
 +
and thus
 +
 +
{{Displayed math||<math>\tan v = \frac{\sin v}{\cos v} = \frac{\dfrac{3}{10}}{-\dfrac{\sqrt{91}}{10}} = -\frac{3}{\sqrt{91}}\,\textrm{.}</math>}}

Current revision

We draw an angle \displaystyle v in the unit circle, and the fact that \displaystyle \sin v = 3/10 means that its y-coordinate equals \displaystyle 3/10.

With the information that is given, we can define a right-angled triangle in the second quadrant which has a hypotenuse of 1 and a vertical side of length 3/10.

We can determine the triangle's remaining side by using the Pythagorean theorem,

\displaystyle a^2 + \Bigl(\frac{3}{10}\Bigr)^2 = 1^2

which gives that

\displaystyle a = \sqrt{1-\Bigl(\frac{3}{10}\Bigr)^2} = \sqrt{1-\frac{9}{100}} = \sqrt{\frac{91}{100}} = \frac{\sqrt{91}}{10}\,\textrm{.}

This means that the angle's x-coordinate is \displaystyle -a, i.e. we have

\displaystyle \cos v=-\frac{\sqrt{91}}{10}

and thus

\displaystyle \tan v = \frac{\sin v}{\cos v} = \frac{\dfrac{3}{10}}{-\dfrac{\sqrt{91}}{10}} = -\frac{3}{\sqrt{91}}\,\textrm{.}
Retrieved from "http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-ImperialCollege/index.php/Solution_4.3:6b"