Question

How do you solve these problems I'm confused

Answer 1

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`

Explanation:

`\displaystyle (OPPOSITE)/(HYPOTENUSE) = sin\:θ \\ (ADJACENT)/(HYPOTENUSE) = cos\:θ \\ (OPPOSITE)/(ADJACENT) = tan\:θ \\ (HYPOTENUSE)/(ADJACENT) = sec\:θ \\ (HYPOTENUSE)/(OPPOSITE) = csc\:θ \\ (ADJACENT)/(OPPOSITE) = cot\:θ`

32.
`\displaystyle cot^(-1)\: (29)/(37) ≈ 51,91122712° ≈ 51,9° \\ \\ OR \\ \\ tan^(-1)\: 1(8)/(29) ≈ 51,91122712° ≈ 51,9°`

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31.
`\displaystyle cot^(-1)\: 1(1)/(4) ≈ 38,65980825° ≈ 38,7° \\ \\ OR \\ \\ tan^(-1)\: (4)/(5) ≈ 38,65980825° ≈ 38,7°`

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30.
`\displaystyle cot^(-1)\: 1(1)/(3) ≈ 36,86989765° ≈ 36,9° \\ \\ OR \\ \\ tan^(-1)\: (3)/(4) ≈ 36,86989765° ≈ 36,9°`

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29.
`\displaystyle sec^(-1)\: 1(2)/(3) ≈ 53,13010235° ≈ 53,1° \\ \\ OR \\ \\ cos^(-1)\: (3)/(5) ≈ 53,13010235° ≈ 53,1°`

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28.
`\displaystyle sec^(-1)\: 1(3)/(5) ≈ 51,31781255° ≈ 51,3° \\ \\ OR \\ \\ cos^(-1)\: (5)/(8) ≈ 51,31781255° ≈ 51,3°`

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27.
`\displaystyle csc^(-1)\: 1(4)/(7) ≈ 39,52119636° ≈ 39,5° \\ \\ OR \\ \\ sin^(-1)\: (7)/(11) ≈ 39,52119636° ≈ 39,5°`

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26.
`\displaystyle cot^(-1)\: (11)/(15) ≈ 53,74616226° ≈ 53,7° \\ \\ OR \\ \\ tan^(-1)\: 1(4)/(11) ≈ 53,74616226° ≈ 53,7°`

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25.
`\displaystyle sec^(-1)\: 2(4)/(9) ≈ 65,85226008° ≈ 65,9° \\ \\ OR \\ \\ cos^(-1)\: (9)/(22) ≈ 65,85226008° ≈ 65,9°`

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24.
`\displaystyle csc^(-1)\: 1(6)/(7) ≈ 32,57897039° ≈ 32,6° \\ \\ OR \\ \\ sin^(-1)\: (7)/(13) ≈ 32,57897039° ≈ 39,5°`

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23.
`\displaystyle cot^(-1)\: (2)/(3) ≈ 56,30993247° ≈ 56,3° \\ \\ OR \\ \\ tan^(-1)\: 1(1)/(2) ≈ 56,30993247° ≈ 56,3°`

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

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