Question #5ddac

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Question #5ddac

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Answer 1 · 2017-11-28T04:14:35

Side a is:

$a = (1650 mm) sin (34^{\circ})$ $a = (1650" mm")sin(34^@)$

$a \approx 922 mm$ $a ~~ 922" mm"$

Side b is:

$b = (1650 mm) cos (34^{\circ})$ $b = (1650" mm")cos(34^@)$

$b \approx 1168 mm$ $b ~~ 1168" mm"$

Answer 2 · 2017-11-28T04:16:36

SOH CAH TOA

Explanation:

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That's based on the information given on the question, now, based on the SOH CAH TOA rules, we can solve for the unknown lengths!

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After solving(assuming you're using degree mode, which is crucial), your side lengths should be approximately $922.7 mm$ $"922.7 mm"$ and $1367.9 mm$ $"1367.9 mm"$ .

Also, as a side note, use Pythagorean's Theorem to confirm your side lengths. $(a^{2} + b^{2} = c^{2})$ $(a^2+b^2=c^2)$

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Question #5ddac

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