Question6 ) In the opposite figure: $\triangle \mathrm{ABC}$ is a right-angled triangle at B , $\tan \theta=\frac{3}{4}$, then $\cos \alpha=$ (a) $\frac{3}{4}$ (b) $-\frac{3}{4}$ (2) $-\frac{4}{5}$ (d) $-\frac{3}{5}$

Studdy Solution
To find $\cos \alpha$, we need the adjacent side to $\alpha$ (which is $AB$) and the hypotenuse $AC$:
$\cos \alpha = \frac{AB}{AC} = \frac{4k}{5k} = \frac{4}{5}$
Since $\alpha$ is the angle at $C$ and considering the orientation of the triangle, $\cos \alpha$ should be negative as it is in the second quadrant:
$\cos \alpha = -\frac{4}{5}$
The value of $\cos \alpha$ is:
$\boxed{-\frac{4}{5}}$