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 $\cos \alpha$, use the cosine definition:
$$\cos \alpha = \frac{\text{adjacent side to } \alpha}{\text{hypotenuse}}$$
The adjacent side to $\alpha$ is $AB = 4$, and the hypotenuse is $AC = 5$. Therefore:
$$\cos \alpha = \frac{4}{5}$$
Since $\alpha$ is in the triangle and the cosine of an angle in a right triangle is positive, the correct answer is:
$$\boxed{\frac{4}{5}}$$