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Math Help - A few quick math questions

  1. #1
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    A few quick math questions

    The triangle is a right triangle.
    'If AB=7cm,AC=10cm,then what is the measure of angle A?'

    'If angle A=38 degrees,AB=10cm,what is the length of BC?'

    'If angle A=20 degrees,AC=18cm
    i)What is the length of the adjecant side?

    ii)What is the length of the opposite side?'

    Any help on these would be fantastic
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  2. #2
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    Quote Originally Posted by karate-kid View Post
    The triangle is a right triangle.
    'If AB=7cm,AC=10cm,then what is the measure of angle A?'

    'If angle A=38 degrees,AB=10cm,what is the length of BC?'

    'If angle A=20 degrees,AC=18cm
    i)What is the length of the adjecant side?

    ii)What is the length of the opposite side?'

    Any help on these would be fantastic
    Q1 You will need to make it clear at which vertex (A, B or C) the right angle is. Alternatively, make it clear which (if any) of the sides mentioned is the hypotenuse.

    Q2 I suppose AC is the hypotenuse. Then \cos 20 = \frac{\text{adjacent}}{18} and \sin 20 = \frac{\text{opposite}}{18}.
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  3. #3
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    For question 1,I think the vertex at which the right angle is is B,and AC is the hypotenuse.
    Thanks for the help
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  4. #4
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    Quote Originally Posted by karate-kid View Post
    The triangle is a right triangle.
    'If AB=7cm,AC=10cm,then what is the measure of angle A?'

    'If angle A=38 degrees,AB=10cm,what is the length of BC?'

    [snip]
    Quote Originally Posted by karate-kid View Post
    For question 1,I think the vertex at which the right angle is is B,and AC is the hypotenuse.
    Thanks for the help
    \cos \theta = \frac{7}{10}. Press some buttons on a calculator to solve for \theta.


    \tan 38^{0} = \frac{BC}{10} (assuming AB is the adjacent side and BC is the opposite side).
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  5. #5
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    BC is actually the adjacent side,and AB is the opposite side
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  6. #6
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    Quote Originally Posted by karate-kid View Post
    BC is actually the adjacent side,and AB is the opposite side
    You should be able to make the required adjustment to what I posted.
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  7. #7
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    Yea I'll figure it out,thanks again
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