# Triangle

Calculate heights of the triangle ABC if sides of the triangle are a=75, b=84 and c=33.

Result

ha =  32.9
hb =  29.4
hc =  74.8

#### Solution:

$\ \\ s = \dfrac{a+b+c}{2} = 96 \ \\ S = \sqrt{ s (s-a)(s-b)(s-c)} = 1234.54 \ \\ \ \\ h _a = \dfrac{ 2S}{a} = \dfrac{ 2 \cdot 1234.54}{ 75} = 32.9$
$h_b = \dfrac{ 2S}{b} = \dfrac{ 2 \cdot 1234.54}{ 84} = 29.4$
$h_c = \dfrac{ 2S}{c} = \dfrac{ 2 \cdot 1234.54}{ 33} = 74.8$

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