Question

k about her sister’s husband?
given. write the letter of the correct choice
13 sin a. \td $\frac{5}{3}$ \th $\frac{3}{5}$
14 cos a
15 tan a \tf $\frac{4}{3}$ \te $\frac{4}{5}$
16 sin b \ti $\frac{3}{sqrt{58}}$ \ta $\frac{3}{7}$
17 cos b
18 tan b \tn $\frac{7}{sqrt{58}}$ \tw $\frac{7}{3}$
19 sin a \tr $\frac{15}{17}$ \tc $\frac{8}{17}$
20 cos a
21 tan a \ts $\frac{17}{8}$ \tl $\frac{8}{15}$
22 sin a \tt $\frac{1}{sqrt{2}}$ \tt $\frac{1}{sqrt{2}}$
23 cos a
24 tan a \tb 1 \tn $sqrt{2}$

Explanation:

Question 13: $\sin A$

Step 1: Recall Sine Definition

In a right triangle, $\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$. For $\angle A$ in $\triangle ABC$ (right-angled at $C$), opposite side to $A$ is $BC = 12$, hypotenuse is $AB = 15$.

Step 2: Calculate $\sin A$

$\sin A = \frac{BC}{AB} = \frac{12}{15} = \frac{4}{5}$? Wait, no, wait. Wait, $BC = 12$, $AC = 9$, $AB = 15$. Wait, opposite to $A$ is $BC$ (length 12), hypotenuse $AB = 15$. Wait, but the options: D is $\frac{5}{3}$, H is $\frac{3}{5}$, F is $\frac{4}{3}$, E is $\frac{4}{5}$. Wait, no, wait, maybe I mixed up. Wait, $\angle A$: adjacent is $AC = 9$, opposite is $BC = 12$, hypotenuse $AB = 15$. Wait, $\sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{12}{15} = \frac{4}{5}$? But E is $\frac{4}{5}$? Wait, no, the options for 13: D $\frac{5}{3}$, H $\frac{3}{5}$, F $\frac{4}{3}$, E $\frac{4}{5}$. Wait, maybe I made a mistake. Wait, $BC = 12$, $AC = 9$, $AB = 15$. So $\sin A = \frac{BC}{AB} = \frac{12}{15} = \frac{4}{5}$, which is E? Wait, no, the problem's 13: $\sin A$, options D $\frac{5}{3}$, H $\frac{3}{5}$, F $\frac{4}{3}$, E $\frac{4}{5}$. Wait, maybe I flipped. Wait, $\angle A$: in triangle $ABC$, right-angled at $C$, so $\angle A$: sides: $AC = 9$ (adjacent), $BC = 12$ (opposite), $AB = 15$ (hypotenuse). So $\sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{12}{15} = \frac{4}{5}$, which is E? But wait, the options for 13: D, H, F, E. Wait, maybe I messed up the triangle. Wait, the triangle has $AC = 9$, $BC = 12$, $AB = 15$. So $\sin A = \frac{BC}{AB} = \frac{12}{15} = \frac{4}{5}$, so E? But the original problem's 13: $\sin A$, options D $\frac{5}{3}$, H $\frac{3}{5}$, F $\frac{4}{3}$, E $\frac{4}{5}$. So E is $\frac{4}{5}$, which is correct. Wait, but maybe I made a mistake. Wait, no, let's check again. $\sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{12}{15} = \frac{4}{5}$, so E. But the options for 13: D, H, F, E. So E is $\frac{4}{5}$, so answer E? Wait, but the user's image: 13 $\sin A$, options D $\frac{5}{3}$, H $\frac{3}{5}$, F $\frac{4}{3}$, E $\frac{4}{5}$. So E is correct.

Question 14: $\cos A$

Step 1: Recall Cosine Definition

$\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}$. For $\angle A$, adjacent side is $AC = 9$, hypotenuse $AB = 15$.

Step 2: Calculate $\cos A$

$\cos A = \frac{AC}{AB} = \frac{9}{15} = \frac{3}{5}$, which is option H.

Question 15: $\tan A$

Step 1: Recall Tangent Definition

$\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$. For $\angle A$, opposite side is $BC = 12$, adjacent side is $AC = 9$.

Step 2: Calculate $\tan A$

$\tan A = \frac{12}{9} = \frac{4}{3}$, which is option F.

Question 16: $\sin B$

Answer:

s:

1. E
2. H
3. F
4. N
5. I
6. W
7. C
8. R
9. L
10. T
11. T
12. B

(Note: Some steps may have had initial confusion, but corrected by re-evaluating the right triangle sides and trigonometric ratios.)