solve right triangle ABC with C the right angle if angle B=56.4 degrees and side c=12.3 Round to the nearest tenth.?

This is one of my final review questions but i do not have a answer key! Just trying to check my work.

I got

Angle A= 33.6

Side a= 6.8

Side b= 10.3

Please provide proof that i am wrong or right

I do not need a simple yes or no! Thanks so much

4 Answers

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  • 2 months ago

    ACB is a right triangle

    in which Angle C = 90°, Angle B = 56.4° and c = 12.3

    Your answer is correct:

    Angle A = 33.6°

    Sides: c = 12.3   b = 10.245   a = 6.807  

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  • 2 months ago

    The sum of the three angles in a triangle will always be 180°.  We have two of the three so we can solve for the third:

    A + B + C = 180

    A + 56.4 + 90 = 180

    A + 146.4 = 180

    A = 33.6°

    Now that we have that we can use the known angles and one known length and the law of sines to solve for the other two unknown lengths:

    a / sin(A) = c / sin(C) and b / sin(B) = c / sin(C)

    a / sin(33.6) = 12.3 / sin(90) and b / sin(56.4) = 12.3 / sin(90)

    a / sin(33.6) = 12.3 / 1 and b / sin(56.4) = 12.3 / 1

    a / sin(33.6) = 12.3 and b / sin(56.4) = 12.3

    a = 12.3 sin(33.6) and b = 12.3 sin(56.4)

    a = 6.8 and b = 10.2

    So other than you not including the label in your angle answer and a little rounding error in your "b", you have the correct answers.

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  • Ash
    Lv 7
    2 months ago

    m∠A = 180 - (m∠B + m∠C) = 180 - (56.4° + 90°) = 33.6°

    ∠C is right angle so side c is hypotenuse

    sinA = a/c

    a = csinA = 12.3 sin33.6 = 6.8

    sinB = b/c

    b = csinB = 12.3 sin56.4 = 10.2

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  • ted s
    Lv 7
    2 months ago

    a = 12.3 sin 33.6°= 6.81 , b = 12.3 sin 56.4 = 10.24...so good work

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