anonymous
  • anonymous
find m∠A.
Mathematics
jamiebookeater
  • jamiebookeater
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jamiebookeater
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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anonymous
  • anonymous
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anonymous
  • anonymous
plz help
taramgrant0543664
  • taramgrant0543664
You can use the formula a^2=c^2+b^2-2abcosA, just put your values in for a,b and c and solve for A

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anonymous
  • anonymous
can you help me plug them in ... im really confused :(
taramgrant0543664
  • taramgrant0543664
So a=4, b=5 and c=2
taramgrant0543664
  • taramgrant0543664
4^2=(5^2)+(2^2)-2(5)(2)cosA I realized in the second portion it should have been bc not ab
taramgrant0543664
  • taramgrant0543664
16=25+4-20cosA 16=29-20cosA -13=-20cosA 0.65=cosA A=? Are you able to finish that?
anonymous
  • anonymous
would divide 0.64 by cosA? if it is right im not really sure how to solve it
taramgrant0543664
  • taramgrant0543664
Nope you have to do the inverse of the cosine function in order to get A by itself so: \[\cos ^{-1}(0.65)\]
anonymous
  • anonymous
ohh... i know im gonna sound kinda dumb.. but how do you do the inverse?
anonymous
  • anonymous
i dont mean to sound mean but can you give me the answer i only have couple of mins left ! im on a timed quiz
taramgrant0543664
  • taramgrant0543664
It's 49.46
anonymous
  • anonymous
thank you so much taramgrant ! really appreciate it :) :)
taramgrant0543664
  • taramgrant0543664
No problem!

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