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Question Number 200785 by cortano12 last updated on 23/Nov/23

Answered by som(math1967) last updated on 24/Nov/23

$$\frac{AB}{sin105}=\frac{AT}{sin45}$$$$\Rightarrow AB=24\times sin75\times \sqrt{2}$$$$\measuredangle ATC=\measuredangle ACT=75$$$$\therefore AT=AC=24cm$$$$△ABC=\frac{1}{2}\times AB\times AC\times sin60$$$$=\frac{1}{2}\times 24\times 24\times sin75\times \sqrt{2}\times \frac{\sqrt{3}}{2}$$$$=144\times \frac{\sqrt{3}+1}{2\sqrt{2}}\times \sqrt{2}{cm}^2$$$$=72\sqrt{3}(\sqrt{3}+1){cm}^2$$$$$$

Commented by cortano12 last updated on 24/Nov/23

$$\mathrm{i}\mathrm{got}\mathrm{the}\mathrm{answer}\mathrm{area}△\mathrm{ABC}$$$$=72\sqrt{3}(1+\sqrt{3})$$

Commented by som(math1967) last updated on 24/Nov/23

$$72\sqrt{3}(1+\sqrt{2})or72\sqrt{3}(1+\sqrt{3})?$$

Commented by cortano12 last updated on 24/Nov/23

$$\mathrm{yes}$$

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