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Question Number 179927 by Tawa11 last updated on 04/Nov/22

Commented by som(math1967) last updated on 04/Nov/22

Commented by som(math1967) last updated on 04/Nov/22

$$3\mathit{b}$$$$\mathrm{\angle }ABX=\mathrm{\angle }AEX=\frac{180-108}{2}=36$$$$sameway\mathrm{\angle }ADE=36$$$$\mathrm{\angle }DEX=108-36=72$$$$\therefore \mathrm{\angle }AXB=\mathrm{\angle }DXE=180-36-72=72$$$$\mathrm{\angle }BAX=108-36=72$$$$\therefore anglesof△ABXare$$$$36,72°,72°$$

Commented by Tawa11 last updated on 04/Nov/22

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}.$$

Answered by som(math1967) last updated on 04/Nov/22

$$ForR48$$$$In△ADBand△BDC$$$$\mathrm{\angle }ADB=\mathrm{\angle }BDC$$$$\mathrm{\angle }CDB=\mathrm{\angle }CBD$$$$\therefore \mathrm{\angle }A=\mathrm{\angle }C$$$$againABCDiscyclic$$$$\therefore \mathrm{\angle }A+\mathrm{\angle }C=180$$$$\Rightarrow 2\mathrm{\angle }A=180$$$$\therefore \mathrm{\angle }A=90$$$$\therefore BDisdiameter$$

Commented by Tawa11 last updated on 04/Nov/22

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}.$$

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