Math / Trigonometry

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

STEP 1

1. $\triangle \mathrm{ABC}$ is a right-angled triangle with the right angle at $B$ .
2. $\tan \theta = \frac{3}{4}$ implies the opposite side to $\theta$ is $3$ and the adjacent side to $\theta$ is $4$ .
3. We need to find $\cos \alpha$ .

STEP 2

1. Use the tangent ratio to find the sides of the triangle.
2. Calculate the hypotenuse of the triangle.
3. Determine $\cos \alpha$ .

STEP 3

Given $\tan \theta = \frac{3}{4}$ , the opposite side to $\theta$ is $3$ and the adjacent side is $4$ . Assign these values to sides $BC$ and $AB$ respectively.

STEP 4

Use the Pythagorean theorem to find the hypotenuse $AC$ . The theorem states:
$AC^2 = AB^2 + BC^2$
Substitute the known values:
$AC^2 = 4^2 + 3^2$ $AC^2 = 16 + 9$ $AC^2 = 25$
Take the square root of both sides:
$AC = \sqrt{25}$ $AC = 5$

STEP 5

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:
$\boxed{\frac{4}{5}}$

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