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Question Number 176014 by cherokeesay last updated on 11/Sep/22

Commented by cherokeesay last updated on 11/Sep/22

$$ThetriangleABCisright-angledatB.$$

Answered by cortano1 last updated on 11/Sep/22

Commented by cherokeesay last updated on 11/Sep/22

$$nicethankyousir.$$

Answered by som(math1967) last updated on 11/Sep/22

$$\frac{OB}{AB}=tan30$$$$\Rightarrow AB=\sqrt{3}\Rightarrow AC=\sqrt{3+2^2}=\sqrt{7}$$$$△ABC\sim △OHC$$$$\therefore \frac{area△OHC}{area△ABC}=\frac{{OC}^2}{{AC}^2}=\frac{1}{7}$$$$\therefore ar.△OHC=\frac{1}{7}△ABC$$$$=\frac{1}{7}\times \frac{1}{2}\times 2\times \sqrt{3}=\frac{\sqrt{3}}{7}squnit$$$$ar.of△AOC=\frac{1}{2}△ABC[OC=OB]$$$$=\frac{\sqrt{3}}{2}squnit$$$$\therefore areaof△AHO=\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{7}$$$$=\frac{5\sqrt{3}}{14}squnit$$$$$$

Commented by cortano1 last updated on 11/Sep/22

$$\mathrm{typo}\mathrm{sir}.\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{7}=\frac{5\sqrt{3}}{14}$$

Commented by som(math1967) last updated on 11/Sep/22

$$yessirthankyou$$

Commented by cherokeesay last updated on 11/Sep/22

$$thankyousir.$$

Answered by mr W last updated on 11/Sep/22

$$AB=\sqrt{3}$$$$AC=\sqrt{7}$$$$\mathrm{\Delta }ABC=\frac{2\times \sqrt{3}}{2}=\sqrt{3}$$$$\mathrm{\Delta }ABO=\frac{1\times \sqrt{3}}{2}=\frac{\sqrt{3}}{2}$$$$\mathrm{\Delta }OHC={\left(\frac{1}{\sqrt{7}}\right)}^2\times \sqrt{3}=\frac{\sqrt{3}}{7}$$$$\Rightarrow \mathrm{\Delta }AHO=\sqrt{3}-\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{7}=\frac{5\sqrt{3}}{14}✓$$

Commented by cortano1 last updated on 11/Sep/22

$$\mathrm{why}\sqrt{5}?$$

Commented by mr W last updated on 11/Sep/22

$$typo$$

Commented by cherokeesay last updated on 11/Sep/22

$$thankyousir.$$

Commented by Tawa11 last updated on 15/Sep/22

$$\mathrm{Great}\mathrm{sirs}$$

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