Question 668543


{{{h= 6sqrt(3)}}}

{{{a}}}= long leg across from {{{60}}}º
{{{b}}} = short leg across from {{{30}}}º
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then {{{a=(1/2)h*sqrt(3)}}}

{{{b=(1/2)h}}} 
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{{{a=(1/2)6sqrt(3)*sqrt(3)}}}

{{{a=(1/2)6(sqrt(3))^2}}}

{{{a=(1/cross(2))cross(6)3*3}}}

{{{a=9}}}



{{{b=(1/2)h}}} 

{{{b=(1/cross(2))cross(6)3sqrt(3)}}}

{{{b=3sqrt(3)}}}

if you plug in value of {{{sqrt(3)=1.73}}}, then you have

{{{h= 6*1.73}}}

{{{highlight(h=10.39)}}}

{{{highlight(a=9)}}}

{{{b=3*1.73}}}

{{{highlight(b=5.19)}}}

now check using Pythagorean theorem:

{{{h^2=a^2+b^2}}}

{{{(10.39)^2=9^2+(5.19)^2}}}


{{{107.95=81+26.94}}}

{{{107.9521=107.94}}}...round to whole numbers

{{{108=108}}}