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In triangle ABC, how do you solve the right triangle given 12, 12 & 20 as sides?

1 Answer

Oct 19, 2015

Largest angle $≅ 1.970 (radians) ≅ {112.9}^{\circ}$ $~=1.970 ("radians") ~= 112.9^@$
Other two angles $≅ 0.586 (radians) ≅ {33.6}^{\circ}$ $~= 0.586 ("radians") ~= 33.6^@$

Explanation:

Note: despite the question statement, a triangle with sides $12, 12, 20$ $12, 12, 20$ is not a right triangle (the sides do not satisfy the Pythagorean Theorem).

The Cosine Law tells us:
$cos (c) = \frac{a^{2} + b^{2} - c^{2}}{2 a b}$ $cos(c) = (a^2+b^2-c^2)/(2ab)$

Setting $c$ $c$ as the longest side ( $20$ $20$ )
we can calculate
$XXX cos (c) = \frac{7}{18}$ $color(white)("XXX")cos(c) = 7/18$

Then use a calculator (or similar) determine
$XXX c = arcsin (\frac{7}{18}) = 1.970$ $color(white)("XXX")c= arcsin(7/18) = 1.970$ (radians)

Similar calculations can be made for $a$ $a$ and $b$ $b$
or
noting that $a = b$ $a=b$ and $a + b + c = π$ $a+b+c=pi$ we can calculate $a$ $a$ and #b" that way.

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