Question
Find the length of $AB.$
✓
Answer
From right $\triangle = ADE$
$\tan 45^{\circ}=\frac{ AE }{ DE } $
$1=\frac{ AE }{30} $
$AE =30 \ cm$
Also, from triangle $DBE$
$\tan 60^{\circ}=\frac{ BE }{ DE }$
$\sqrt{3}=\frac{ BE }{30}$
$BE =30 \sqrt{3} \ cm$
Therefore $AB=AE+BE=30+30 \sqrt{3}$
$=30(1+\sqrt{3}) \ cm$
$\tan 45^{\circ}=\frac{ AE }{ DE } $
$1=\frac{ AE }{30} $
$AE =30 \ cm$
Also, from triangle $DBE$
$\tan 60^{\circ}=\frac{ BE }{ DE }$
$\sqrt{3}=\frac{ BE }{30}$
$BE =30 \sqrt{3} \ cm$
Therefore $AB=AE+BE=30+30 \sqrt{3}$
$=30(1+\sqrt{3}) \ cm$
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