|
Example 1
What is $\sin x$ if $2 \tan x + 2 = 5$ and $0 \lt x \lt \dfrac{\pi}{2}$
-
$\dfrac{2}{\sqrt {13}}$
-
$\dfrac{3}{\sqrt {13}}$
-
$-\dfrac{3}{\sqrt {13}}$
-
$\dfrac{2}{3}$
-
$\dfrac{4}{\sqrt {13}}$
Solution
- We first use the given equation to find $\tan \theta$.
$2 \tan x + 2 = 5$
$ 2\tan x = 5-2$
$ \tan x = \dfrac{3}{2} $
- We now use the definition of $\tan x$ using a point with coordinates $(a,b)$ on the terminal side of $x$
$\tan x = \dfrac{3}{2} = \dfrac{b}{a}$
- Since $x$ is in quadrant $I$, $a$ and $b$ are positive. Hence
$a = 2$ and $b = 3$
- Distance r from the origin to the point with coordinates (a,b) is given by
$r = \sqrt{a^2+b^2}=\sqrt{4+9}=\sqrt{13}$
$\sin x= \dfrac{b}{r} = \dfrac{3}{\sqrt {13}}$
Answer B
|