Question

match the following
(there is a right triangle image with right angle at b, sides labeled: ab, bc, ac; and angle labels a, b, c)
tan a
tan b
sin a
cos a
size of angle b
90°
\frac{15}{8}
\frac{8}{15}
\frac{15}{17}
\frac{8}{17}

Explanation:

Step1: Analyze the right triangle

We have a right triangle \( \triangle ABC \) with \( \angle B = 90^\circ \), \( AB = 8 \), \( BC = 15 \), and \( AC = 17 \) (by Pythagorean theorem: \( 8^2 + 15^2 = 64 + 225 = 289 = 17^2 \)).

Step2: Calculate \( \cos A \)

In a right triangle, \( \cos A=\frac{\text{adjacent to }A}{\text{hypotenuse}} \). The adjacent side to \( \angle A \) is \( AB = 8 \)? Wait, no, wait. Wait, \( \angle A \): the sides: opposite to \( A \) is \( BC = 15 \), adjacent is \( AB = 8 \), hypotenuse \( AC = 17 \). Wait, no, \( \cos A=\frac{AB}{AC}=\frac{8}{17} \)? Wait, no, maybe I mixed up. Wait, let's re - identify:

In \( \triangle ABC \), right - angled at \( B \):

- For \( \angle A \):
- Opposite side (\( \text{opp} \)): \( BC = 15 \)
- Adjacent side (\( \text{adj} \)): \( AB = 8 \)
- Hypotenuse (\( \text{hyp} \)): \( AC = 17 \)

- \( \cos A=\frac{\text{adj}}{\text{hyp}}=\frac{AB}{AC}=\frac{8}{17} \)? Wait, but the options have \( \frac{15}{17} \) and \( \frac{8}{17} \). Wait, maybe I made a mistake. Wait, no, let's check \( \sin A \): \( \sin A=\frac{\text{opp}}{\text{hyp}}=\frac{BC}{AC}=\frac{15}{17} \)

- \( \tan A=\frac{\text{opp}}{\text{adj}}=\frac{BC}{AB}=\frac{15}{8} \)

- For \( \angle B = 90^\circ \), \( \sin B = 1 \), but we have \( \sin B \) to match? Wait, the "Size of Angle B" is \( 90^\circ \), that's correct.

- For \( \tan C \): In \( \triangle ABC \), \( \angle C \): opposite side is \( AB = 8 \), adjacent side is \( BC = 15 \), so \( \tan C=\frac{\text{opp to }C}{\text{adj to }C}=\frac{AB}{BC}=\frac{8}{15} \)? Wait, no, wait \( \angle C \): opposite side is \( AB = 8 \), adjacent side is \( BC = 15 \), so \( \tan C=\frac{AB}{BC}=\frac{8}{15} \)? But the options have \( \frac{8}{15} \) and \( \frac{15}{8} \). Wait, maybe I mixed up \( \angle A \) and \( \angle C \).

Wait, let's re - assign:

- \( \cos A \): adjacent to \( A \) is \( AB = 8 \), hypotenuse \( AC = 17 \), so \( \cos A=\frac{8}{17} \)? No, the option for \( \cos A \) is either \( \frac{15}{17} \) or \( \frac{8}{17} \). Wait, maybe I got the adjacent and opposite wrong. Wait, \( AB = 8 \), \( BC = 15 \), \( AC = 17 \). So \( \angle A \): the sides: \( AB \) is adjacent, \( BC \) is opposite, \( AC \) is hypotenuse. So \( \cos A=\frac{AB}{AC}=\frac{8}{17} \), \( \sin A=\frac{BC}{AC}=\frac{15}{17} \), \( \tan A=\frac{BC}{AB}=\frac{15}{8} \)

- \( \sin B \): since \( \angle B = 90^\circ \), \( \sin B = 1 \), but the option for "Size of Angle B" is \( 90^\circ \), which is correct.

- \( \tan C \): \( \angle C \): opposite side is \( AB = 8 \), adjacent side is \( BC = 15 \), so \( \tan C=\frac{AB}{BC}=\frac{8}{15} \)

Let's match each term:

1. \( \cos A \): \( \frac{8}{17} \)
2. \( \tan A \): \( \frac{15}{8} \)
3. \( \sin A \): \( \frac{15}{17} \)
4. \( \sin B \): Since \( \angle B = 90^\circ \), \( \sin B = 1 \), but the option for "Size of Angle B" is \( 90^\circ \), so "Size of Angle B" matches \( 90^\circ \)
5. \( \tan C \): \( \frac{8}{15} \)

Wait, let's do the matching properly:

- \( \cos A \): adjacent over hypotenuse \(=\frac{8}{17}\)
- \( \tan A \): opposite over adjacent \(=\frac{15}{8}\)
- \( \sin A \): opposite over hypotenuse \(=\frac{15}{17}\)
- "Size of Angle B": \( 90^\circ \) (since \( \angle B \) is the right angle)
- \( \tan C \): For \( \angle C \), opposite side is \( AB = 8 \), adjacent side is \( BC = 15 \), so \( \tan C=\frac{8}{15}\)

So the matches are:

- \( \cos A \): \( \frac{8}{17} \)
- \( \tan A \): \( \frac{15}{8} \)
- \( \sin A \): \( \frac{15}{17} \)
- "Size of Angle B": \( 90^\circ \)
- \( \tan C \): \( \frac{8}{15} \)

Let's check the original problem's left - hand side and right - hand side:

Left - hand side (LHS) items: \( \cos A \), \( \tan A \), \( \sin A \), \( \sin B \) (Size of Angle B), \( \tan C \)

Right - hand side (RHS) items: \( 90^\circ \), \( \frac{15}{8} \), \( \frac{8}{15} \), \( \frac{15}{17} \), \( \frac{8}{17} \)

So:

- \( \cos A \): \( \frac{8}{17} \)
- \( \tan A \): \( \frac{15}{8} \)
- \( \sin A \): \( \frac{15}{17} \)
- Size of Angle B: \( 90^\circ \)
- \( \tan C \): \( \frac{8}{15} \)

If we assume that the question is to match each of the left - side terms to the right - side terms, here are the correct matches:

1. \( \cos A \) - \( \frac{8}{17} \)
2. \( \tan A \) - \( \frac{15}{8} \)
3. \( \sin A \) - \( \frac{15}{17} \)
4. Size of Angle B - \( 90^\circ \)
5. \( \tan C \) - \( \frac{8}{15} \)

But since the user might be asking for a specific match, let's take an example. Let's say we want to find \( \sin A \). \( \sin A=\frac{\text{opposite to }A}{\text{hypotenuse}}=\frac{15}{17} \)

Answer:

If we take \( \sin A \), the answer is \( \frac{15}{17} \); if we take \( \tan A \), the answer is \( \frac{15}{8} \); if we take \( \cos A \), the answer is \( \frac{8}{17} \); if we take "Size of Angle B", the answer is \( 90^\circ \); if we take \( \tan C \), the answer is \( \frac{8}{15} \)

(Assuming the question is to match each term, and if we pick \( \sin A \) as an example, the answer is \( \frac{15}{17} \))