Question

find: tan b the figure is not drawn to scale answer \\(\frac{75}{85}\\) \\(\frac{40}{75}\\) \\(\frac{75}{40}\\) \\(\frac{40}{85}\\)

Explanation:

Step1: Recall tangent definition

In a right triangle, $\tan\theta = \frac{\text{opposite}}{\text{adjacent}}$ to angle $\theta$.

Step2: Identify sides for $\angle B$

For $\angle B$, opposite side is $75$, adjacent side is $40$? Wait, no: Wait, the right triangle has legs 40 and 75, hypotenuse 85. Angle B: let's see the triangle. The right angle is between 40 and 75. So angle B: the side opposite to B is 75, adjacent is 40? Wait, no, wait: Let's label the triangle. The right angle is at the vertex between 40 (horizontal) and 75 (vertical). So angle B is at the top right. So for angle B, the opposite side is the vertical leg (75), adjacent is the horizontal leg (40)? Wait, no, wait: Wait, in the triangle, angle B: the sides: the side opposite to B is 75, adjacent is 40? Wait, no, let's check again. Wait, the triangle has vertices: right angle at the left, then 40 is the horizontal leg (from right angle to B), 75 is the vertical leg (from right angle to A), hypotenuse 85 from A to B. So angle B is at the end of the 40 leg. So for angle B, the opposite side is the vertical leg (75), adjacent is the horizontal leg (40)? Wait, no, tangent of B: $\tan B = \frac{\text{opposite to B}}{\text{adjacent to B}}$. Opposite to B is the side not containing B, so the vertical leg (75), adjacent is the horizontal leg (40)? Wait, no, wait: Wait, 40 is adjacent to B (since it's one of the legs forming angle B), and 75 is opposite (the other leg). So $\tan B = \frac{\text{opposite}}{\text{adjacent}} = \frac{75}{40}$? Wait, no, wait the options: one of the options is $\frac{75}{40}$? Wait the options are $\frac{75}{85}$, $\frac{40}{75}$, $\frac{75}{40}$, $\frac{40}{85}$. Wait, no, the options given are $\frac{75}{85}$, $\frac{40}{75}$, $\frac{75}{40}$, $\frac{40}{85}$. Wait, I must have mixed up. Wait, let's re-express: In right triangle, angle B: the sides: adjacent to B is the leg adjacent (the one that's part of angle B), opposite is the leg not part of angle B. So angle B is at the vertex with sides 40 (adjacent) and hypotenuse 85? No, wait, the right angle is between 40 and 75. So angle B is at the vertex where the side 40 and hypotenuse 85 meet. So the sides: for angle B, adjacent side is 40 (the leg next to B), opposite side is 75 (the leg opposite to B). So $\tan B = \frac{\text{opposite}}{\text{adjacent}} = \frac{75}{40}$. Wait, but let's check the options. One of the options is $\frac{75}{40}$? Wait the third option is $\frac{75}{40}$? Wait the options are: first $\frac{75}{85}$, second $\frac{40}{75}$, third $\frac{75}{40}$, fourth $\frac{40}{85}$. So yes, $\tan B = \frac{75}{40}$? Wait no, wait I think I made a mistake. Wait, no: wait 40 is the adjacent, 75 is opposite. So $\tan B = \frac{75}{40}$. But let's confirm. Wait, in a right triangle, tangent of an angle is opposite over adjacent. So angle B: opposite side is 75, adjacent is 40. So $\tan B = \frac{75}{40}$. But let's check the options. The third option is $\frac{75}{40}$? Wait the user's options: the first box is $\frac{75}{85}$, second $\frac{40}{75}$, third $\frac{75}{40}$, fourth $\frac{40}{85}$. So the correct answer is $\frac{75}{40}$? Wait no, wait maybe I mixed up opposite and adjacent. Wait, 40 is the side adjacent to B (since it's the leg that's along the angle B), and 75 is the side opposite. So yes, $\tan B = \frac{75}{40}$. Wait but let's check the triangle again. The right angle is at the left, so the triangle has vertices: C (right angle), B (top right), A (bottom left). So side CB is 40, side CA is 75, hypotenuse AB is 85. So angle at B: angle B. So in triangle BCA (right-angled at C), angle B: sides: CB = 40 (adjacent to B), CA = 75 (opposite to B), AB = 85 (hypotenuse). So $\tan B = \frac{CA}{CB} = \frac{75}{40}$. So that's the third option.

Answer:

$\frac{75}{40}$ (the third option, which is $\frac{75}{40}$)