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Question-200785




Question Number 200785 by cortano12 last updated on 23/Nov/23
Answered by som(math1967) last updated on 24/Nov/23
 ((AB)/(sin105))=((AT)/(sin45))  ⇒AB=24×sin75×(√2)   ∡ATC=∡ACT=75  ∴AT=AC=24cm  △ABC=(1/2)×AB×AC×sin60  =(1/2)×24×24×sin75×(√2)×((√3)/2)  =144×(((√3)+1)/(2(√2)))×(√2)cm^2   =72(√3)((√3)+1)cm^2
$$\:\frac{{AB}}{{sin}\mathrm{105}}=\frac{{AT}}{{sin}\mathrm{45}} \\ $$$$\Rightarrow{AB}=\mathrm{24}×{sin}\mathrm{75}×\sqrt{\mathrm{2}} \\ $$$$\:\measuredangle{ATC}=\measuredangle{ACT}=\mathrm{75} \\ $$$$\therefore{AT}={AC}=\mathrm{24}{cm} \\ $$$$\bigtriangleup{ABC}=\frac{\mathrm{1}}{\mathrm{2}}×{AB}×{AC}×{sin}\mathrm{60} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{24}×\mathrm{24}×{sin}\mathrm{75}×\sqrt{\mathrm{2}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$=\mathrm{144}×\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}×\sqrt{\mathrm{2}}{cm}^{\mathrm{2}} \\ $$$$=\mathrm{72}\sqrt{\mathrm{3}}\left(\sqrt{\mathrm{3}}+\mathrm{1}\right){cm}^{\mathrm{2}} \\ $$$$ \\ $$
Commented by cortano12 last updated on 24/Nov/23
i got the answer area △ABC   = 72(√3)(1+(√3) )
$$\mathrm{i}\:\mathrm{got}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{area}\:\bigtriangleup\mathrm{ABC} \\ $$$$\:=\:\mathrm{72}\sqrt{\mathrm{3}}\left(\mathrm{1}+\sqrt{\mathrm{3}}\:\right) \\ $$
Commented by som(math1967) last updated on 24/Nov/23
72(√3)(1+(√2)) or 72(√3)(1+(√3))?
$$\mathrm{72}\sqrt{\mathrm{3}}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\:{or}\:\mathrm{72}\sqrt{\mathrm{3}}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)? \\ $$
Commented by cortano12 last updated on 24/Nov/23
yes
$$\mathrm{yes} \\ $$

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