Question

given right triangle mno, which represents the value of (cos(40^circ))? the triangle has a right angle at n, angle at o is (40^circ), angle at m is (50^circ). the options are (\frac{on}{mo}), (\frac{mn}{mo}), (\frac{mn}{on}), (\frac{on}{mn}) (exact labels may vary due to image clarity but the context is trigonometric ratios in a right triangle with angle (40^circ) at o).

Explanation:

Step1: Recall cosine definition

In a right triangle, $\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}$. For $\angle M$, adjacent side is $MN$, hypotenuse is $MO$.

Step2: Identify sides for $\angle M$

In $\triangle MNO$ (right-angled at $N$), $\angle M = 50^\circ$. Adjacent to $\angle M$ is $MN$, hypotenuse is $MO$. So $\cos(M) = \frac{MN}{MO}$.

Answer:

$\frac{MN}{MO}$ (corresponding to the option with $\frac{MN}{MO}$)