Question

if cosx° = 3/5, what is the value of b? b = 4 b = 5 b = 6 b = 7

Explanation:

Step1: Recall cosine definition

In a right - triangle, $\cos x=\frac{\text{adjacent}}{\text{hypotenuse}}$. Given $\cos x = \frac{3}{5}$, assume the adjacent side to angle $x$ is $3$ and the hypotenuse is $5$.

Step2: Apply Pythagorean theorem

Let the side we want to find be $b$. By the Pythagorean theorem $a^{2}+b^{2}=c^{2}$, where $c = 5$ (hypotenuse) and $a = 3$ (adjacent side). So $3^{2}+b^{2}=5^{2}$, which simplifies to $9 + b^{2}=25$. Then $b^{2}=25 - 9=16$.

Step3: Solve for $b$

Taking the square root of both sides, $b=\sqrt{16}=4$ (we take the positive value since $b$ represents the length of a side of a triangle).

Answer:

$b = 4$