Question

Help me 2 write my homework.....Pls...​

Answer 1

Answer:

in the ∆ABC

angle A=60°

using

$\tan(60 ) = \frac{BC}{AB} \ \\ \\ \tan(60) = \frac{BC}{2 \sqrt{3} } \\ \\ \sqrt{3} = \frac{BC}{2 \sqrt{3} } \\ \\ \sqrt{3} (2 \sqrt{3} ) = BC \\ \\ BC = 2(3) \\ \\ BC = 6cm$

now,

we know that

area of triangle=base×height/2

in the∆ABC

base=BC=6

height=AB=2√3

Substitute in the above formula

$area \: of \: triangle = \frac{6(2 \sqrt{3}) }{2} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 6 \sqrt{3}$

${\mathrm{\green{\underline{area \: of \: triangle=6 \sqrt{3} {cm}^{2} }}}}$

Answer 2

Answer:

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