Question

How do you solve these problems I'm confused

Answer 1

Answer:

32. $\displaystyle 51,9° ≈ m∠A$

31. $\displaystyle 38,7° ≈ m∠A$

30. $\displaystyle 36,9° ≈ m∠A$

29. $\displaystyle 53,1° ≈ m∠A$

28. $\displaystyle 51,3° ≈ m∠A$

27. $\displaystyle 39,5° ≈ m∠A$

26. $\displaystyle 53,7° ≈ m∠A$

25. $\displaystyle 65,9° ≈ m∠A$

24. $\displaystyle 32,6° ≈ m∠A$

23. $\displaystyle 56,3° ≈ m∠A$

Step-by-step explanation:

$\displaystyle \frac{OPPOSITE}{HYPOTENUSE} = sin\:θ \\ \frac{ADJACENT}{HYPOTENUSE} = cos\:θ \\ \frac{OPPOSITE}{ADJACENT} = tan\:θ \\ \frac{HYPOTENUSE}{ADJACENT} = sec\:θ \\ \frac{HYPOTENUSE}{OPPOSITE} = csc\:θ \\ \frac{ADJACENT}{OPPOSITE} = cot\:θ$

32. $\displaystyle cot^{-1}\: \frac{29}{37} ≈ 51,91122712° ≈ 51,9° \\ \\ OR \\ \\ tan^{-1}\: 1\frac{8}{29} ≈ 51,91122712° ≈ 51,9°$

__________________________________________________________

31. $\displaystyle cot^{-1}\: 1\frac{1}{4} ≈ 38,65980825° ≈ 38,7° \\ \\ OR \\ \\ tan^{-1}\: \frac{4}{5} ≈ 38,65980825° ≈ 38,7°$

__________________________________________________________

30. $\displaystyle cot^{-1}\: 1\frac{1}{3} ≈ 36,86989765° ≈ 36,9° \\ \\ OR \\ \\ tan^{-1}\: \frac{3}{4} ≈ 36,86989765° ≈ 36,9°$

__________________________________________________________

29. $\displaystyle sec^{-1}\: 1\frac{2}{3} ≈ 53,13010235° ≈ 53,1° \\ \\ OR \\ \\ cos^{-1}\: \frac{3}{5} ≈ 53,13010235° ≈ 53,1°$

__________________________________________________________

28. $\displaystyle sec^{-1}\: 1\frac{3}{5} ≈ 51,31781255° ≈ 51,3° \\ \\ OR \\ \\ cos^{-1}\: \frac{5}{8} ≈ 51,31781255° ≈ 51,3°$

__________________________________________________________

27. $\displaystyle csc^{-1}\: 1\frac{4}{7} ≈ 39,52119636° ≈ 39,5° \\ \\ OR \\ \\ sin^{-1}\: \frac{7}{11} ≈ 39,52119636° ≈ 39,5°$

__________________________________________________________

26. $\displaystyle cot^{-1}\: \frac{11}{15} ≈ 53,74616226° ≈ 53,7° \\ \\ OR \\ \\ tan^{-1}\: 1\frac{4}{11} ≈ 53,74616226° ≈ 53,7°$

__________________________________________________________

25. $\displaystyle sec^{-1}\: 2\frac{4}{9} ≈ 65,85226008° ≈ 65,9° \\ \\ OR \\ \\ cos^{-1}\: \frac{9}{22} ≈ 65,85226008° ≈ 65,9°$

__________________________________________________________

24. $\displaystyle csc^{-1}\: 1\frac{6}{7} ≈ 32,57897039° ≈ 32,6° \\ \\ OR \\ \\ sin^{-1}\: \frac{7}{13} ≈ 32,57897039° ≈ 39,5°$

__________________________________________________________

23. $\displaystyle cot^{-1}\: \frac{2}{3} ≈ 56,30993247° ≈ 56,3° \\ \\ OR \\ \\ tan^{-1}\: 1\frac{1}{2} ≈ 56,30993247° ≈ 56,3°$

* Whenever you are solving for angle measures inside right triangles, ALWAYS use the inverse trigonometric ratios.

I am delighted to assist you anytime.