Question

timed problem
score: 7/10
current time: 303.1
find: cos a
the figure is not drawn to scale.
right triangle with leg 51 (adjacent to angle a), leg 68, hypotenuse 85
answer options: \\(\frac{68}{85}\\), \\(\frac{51}{85}\\), \\(\frac{68}{51}\\), \\(\frac{51}{68}\\)

Explanation:

Step1: Recall cosine definition

In a right triangle, $\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}$. For angle \( A \), adjacent side is \( 51 \), hypotenuse is \( 85 \).

Step2: Calculate $\cos A$

$\cos A = \frac{51}{85}$? Wait, no, wait. Wait, adjacent to angle \( A \): wait, the right triangle has legs \( 51 \) and \( 68 \), hypotenuse \( 85 \). Wait, angle \( A \): the adjacent side to \( A \) is \( 51 \)? Wait, no, wait. Wait, in the right triangle, angle \( A \) is at the vertex with leg \( 51 \) and hypotenuse \( 85 \). Wait, no, adjacent side to angle \( A \) is the leg adjacent, so adjacent is \( 51 \), opposite is \( 68 \), hypotenuse \( 85 \). Wait, no, wait: $\cos A = \frac{\text{adjacent to } A}{\text{hypotenuse}}$. Adjacent to \( A \) is \( 51 \)? Wait, no, wait the legs: one leg is \( 51 \), another is \( 68 \), hypotenuse \( 85 \). So angle \( A \): the sides: adjacent is \( 51 \), hypotenuse \( 85 \)? Wait, no, wait, maybe I mixed up. Wait, let's label the triangle: right angle at the vertex between \( 51 \) and \( 68 \), so angle \( A \) is at the vertex with side \( 51 \), so the sides: adjacent to \( A \) is \( 51 \), opposite is \( 68 \), hypotenuse \( 85 \). Wait, but then $\cos A = \frac{51}{85}$? But wait, the options: let's check the options. Wait, the options are $\frac{68}{85}$, $\frac{51}{85}$, $\frac{68}{51}$, $\frac{51}{68}$. Wait, maybe I made a mistake. Wait, adjacent to angle \( A \): maybe the adjacent side is \( 51 \), hypotenuse \( 85 \)? No, wait, no: in a right triangle, for angle \( A \), the adjacent side is the leg that is part of angle \( A \) (other than the hypotenuse). So angle \( A \) is formed by the leg \( 51 \) and the hypotenuse \( 85 \). So the adjacent side is \( 51 \), hypotenuse \( 85 \). Wait, but \( \frac{51}{85} \) simplifies to \( \frac{3}{5} \), but let's check the options. Wait, the options include \( \frac{51}{85} \)? Wait, the second option is \( \frac{51}{85} \)? Wait, no, the user's image: the options are \( \frac{68}{85} \), \( \frac{51}{85} \), \( \frac{68}{51} \), \( \frac{51}{68} \). Wait, maybe I mixed up adjacent and opposite. Wait, angle \( A \): the other leg is \( 68 \), so maybe adjacent is \( 51 \), opposite is \( 68 \), hypotenuse \( 85 \). So $\cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{51}{85}$? Wait, but let's recalculate. Wait, \( 51^2 + 68^2 = 2601 + 4624 = 7225 \), and \( 85^2 = 7225 \), so that's correct. So $\cos A = \frac{51}{85}$? Wait, no, wait: adjacent to angle \( A \) is the side next to angle \( A \), which is \( 51 \), and hypotenuse is \( 85 \). So $\cos A = \frac{51}{85}$? But wait, the first option is \( \frac{68}{85} \), which is \( \frac{4}{5} \), and \( 68 \) is the opposite side to angle \( A \). Wait, maybe I got the angle wrong. Wait, angle \( B \) would have adjacent \( 68 \), hypotenuse \( 85 \), so $\cos B = \frac{68}{85}$. But the question is $\cos A$. So angle \( A \): adjacent is \( 51 \), hypotenuse \( 85 \), so $\cos A = \frac{51}{85}$. Wait, but let's check the options. The second option is \( \frac{51}{85} \), so that's the answer? Wait, no, wait, maybe I made a mistake. Wait, let's re-express: in a right triangle, cosine of an angle is adjacent over hypotenuse. So for angle \( A \), the sides: the two legs are \( 51 \) (vertical) and \( 68 \) (horizontal), hypotenuse \( 85 \). So angle \( A \) is at the bottom left, so the adjacent side is \( 51 \) (vertical), hypotenuse \( 85 \), so $\cos A = \frac{51}{85}$. Yes, that's correct. So the answer is \( \frac{51}{85} \), which is the second optio…

Answer:

$\frac{51}{85}$ (corresponding to the option with $\frac{51}{85}$)