Question

what is cot(∠b)? reduce fractional answers to lowest terms. diagram: right triangle with right angle at c, ac=24, cb=7, ab=25, vertices a, b, c

Explanation:

Step1: Recall the definition of cotangent

In a right triangle, the cotangent of an acute angle is the ratio of the adjacent side to the opposite side. For $\angle B$, we need to identify the adjacent and opposite sides.

Step2: Identify sides relative to $\angle B$

In right triangle $ABC$ with right angle at $C$, for $\angle B$:
- The adjacent side (adjacent to $\angle B$) is $BC = 7$.
- The opposite side (opposite to $\angle B$) is $AC = 24$.

Step3: Calculate $\cot(\angle B)$

Using the definition $\cot(\theta)=\frac{\text{adjacent}}{\text{opposite}}$, we substitute the values:
$\cot(\angle B)=\frac{BC}{AC}=\frac{7}{24}$? Wait, no, wait. Wait, in right triangle, angle at C is right angle, so sides:
Wait, angle B: the sides: adjacent is BC (length 7), opposite is AC (length 24)? Wait, no, wait. Wait, cotangent is adjacent over opposite. Wait, let's re - check. In right triangle $ABC$, right - angled at $C$. So $\angle B$: the sides:
- Adjacent side to $\angle B$: the side that is part of $\angle B$ and not the hypotenuse. So $BC$ is adjacent (length 7), and the opposite side is $AC$ (length 24)? Wait, no, wait. Wait, $\cot(\theta)=\frac{\cos(\theta)}{\sin(\theta)}$, and $\cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}$, $\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}$, so $\cot(\theta)=\frac{\text{adjacent}}{\text{opposite}}$.
Wait, for $\angle B$:
- Adjacent side: $BC = 7$ (because it is adjacent to $\angle B$ and forms the angle with the hypotenuse)
- Opposite side: $AC = 24$ (because it is opposite to $\angle B$)
Wait, no, wait, I think I made a mistake. Wait, in triangle $ABC$, right - angled at $C$, so sides:
- $AC = 24$ (vertical leg)
- $BC = 7$ (horizontal leg)
- $AB = 25$ (hypotenuse)
For $\angle B$:
- The adjacent side (the leg that is adjacent to $\angle B$) is $BC = 7$? No, wait, no. Wait, $\angle B$ is at vertex $B$, so the two sides forming $\angle B$ are $BC$ and $AB$. The side opposite to $\angle B$ is $AC$, and the side adjacent to $\angle B$ (other than the hypotenuse) is $BC$? Wait, no, the adjacent side should be the leg that is part of $\angle B$ and is not the hypotenuse. Wait, $\angle B$ is between $BC$ and $AB$. So the adjacent side (leg) is $BC$, and the opposite side (leg) is $AC$.
So $\cot(\angle B)=\frac{\text{adjacent}}{\text{opposite}}=\frac{BC}{AC}=\frac{7}{24}$? Wait, no, wait, that can't be. Wait, no, wait, I think I mixed up adjacent and opposite. Wait, let's use SOH - CAH - TOA.
- SOH: $\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}$
- CAH: $\cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}$
- TOA: $\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}$
And $\cot(\theta)=\frac{1}{\tan(\theta)}=\frac{\text{adjacent}}{\text{opposite}}$
For $\angle B$:
- Opposite side: $AC = 24$ (the side opposite to $\angle B$)
- Adjacent side: $BC = 7$ (the side adjacent to $\angle B$, which is not the hypotenuse)
Wait, no, wait, in the right triangle at $C$, $\angle B$: the sides:
- The side opposite $\angle B$ is $AC$ (length 24)
- The side adjacent to $\angle B$ is $BC$ (length 7)
So $\cot(\angle B)=\frac{\text{adjacent}}{\text{opposite}}=\frac{BC}{AC}=\frac{7}{24}$? Wait, but let's check with the triangle sides. Wait, $AC = 24$, $BC = 7$, $AB = 25$. Let's verify with Pythagoras: $7^{2}+24^{2}=49 + 576=625=25^{2}$, so the triangle is valid.
Wait, but maybe I had adjacent and opposite reversed. Wait, $\angle B$: the angle at $B$, so the sides:
- The leg adjacent to $\angle B$ is $BC$ (because it is next to $\angle B$), and the leg opposite is $AC$. So $\cot(\angle B)=\frac{BC}{AC}=\frac{7}{24}$? Wait, no, wait, no. Wait, $\cot(\theta)=\frac{\text{adjacent}}{\text{opposite}}$. Wait, if $\theta=\angle B$, then the adjacent side is the side that is part of $\theta$ and is not the hypotenuse. So in $\angle B$, the two sides are $BC$ (length 7) and $AB$ (hypotenuse, length 25). So the adjacent side (leg) is $BC$, and the opposite side (leg) is $AC$ (length 24). So $\cot(\angle B)=\frac{BC}{AC}=\frac{7}{24}$? Wait, that seems correct. Wait, but let's think again. Suppose we have angle $B$, then:
- $\tan(\angle B)=\frac{\text{opposite}}{\text{adjacent}}=\frac{AC}{BC}=\frac{24}{7}$, so $\cot(\angle B)=\frac{1}{\tan(\angle B)}=\frac{7}{24}$. Yes, that's correct.

Answer:

$\frac{7}{24}$