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Um,What ?
is it b? :)
Yeah right :)
great thanks! :)
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You have to use, \[\sin \theta =P/H\] \[\sin 30^0=P/2\sqrt{3}\] \[P=\sin30^0*2 \sqrt{3}\] \[P=\frac{ 1 }{ 2 } \times 2\sqrt{3}\] \[P=\sqrt{3}\]
No. The sides of a 30-60-90 triangle are of proportions 1:sqrt3:2. If we apply that proportion to the given triangle, the sides are sqrt3, 3, 2sqrt3.
That is the same as (1:sqrt3:2)*sqrt3
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