How do you solve the right triangle ABC given b=2, A=8?

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Answer 1 · 2018-05-03T15:26:35

$c = 2 \sqrt{17} \approx 8.25$ $c = 2 sqrt 17 approx 8.25$ cm

$a^{2} + b^{2} = c^{2}$ $a^2 + b^2 = c^2$

In which c is always the longest line in the triangle which is the hypotenuse of the triangle.

Assuming that the A and b that you stated are the opposite and the adjacent, we can substitute it in the formula.

Substitution

$8^{2} + 2^{2} = c^{2}$ $8^2 + 2^2 = c^2$

This gives you:

$c^{2} = 68$ $c^2 = 68$

To solve for c,

$c = \sqrt{68} = 2 \sqrt{17}$ $c = sqrt68 = 2 sqrt 17$

$c \approx 8.25$ $c approx 8.25$ cm

If angles are provided, you can use the sine, cosine or tangent rule.