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Question Number 79361 by TawaTawa last updated on 24/Jan/20

Commented by mr W last updated on 24/Jan/20

$$inthefigurepointOisnotdisplayed.$$$$weassumeOistheintersectionofAD$$$$andEC.$$

Answered by mr W last updated on 24/Jan/20

Commented by mr W last updated on 24/Jan/20

$$\alpha =180-30-40=110°$$$$\beta =\alpha =110°$$$$\gamma =\frac{\beta }{2}=55°$$$$\delta +\gamma =180°$$$$\Rightarrow \mathrm{\angle }ABC=\delta =180-\gamma =180-55=125°$$$$$$$$weseeitisevennotnecessarythat$$$$EDtangentsthecircleasshown.$$

Commented by TawaTawa last updated on 24/Jan/20

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

Commented by mr W last updated on 24/Jan/20

$$isthissolutioncorrect?$$$$canyouunderstandmysolution?$$

Commented by TawaTawa last updated on 24/Jan/20

$$\mathrm{I}\mathrm{understand}\mathrm{your}\mathrm{solution}\mathrm{sir}$$

Commented by TawaTawa last updated on 24/Jan/20

$$\alpha +30+40=180\left(\mathrm{angle}\mathrm{in}\mathrm{\Delta }\right)$$$$\alpha =\beta \left(\mathrm{vertically}\mathrm{opposite}\mathrm{angle}\right)$$$$\gamma +\delta =180\left(\mathrm{opposite}\mathrm{angle}\mathrm{in}\mathrm{cyclic}\mathrm{quadrillaterals}\right)$$$$$$$$\mathrm{I}\mathrm{will}\mathrm{use}\mathrm{your}\mathrm{workings}\mathrm{sir}.\mathrm{I}\mathrm{understand}.$$

Commented by mr W last updated on 24/Jan/20

$$fine!thenyoucansolveQ79368by$$$$yourself.$$

Commented by TawaTawa last updated on 24/Jan/20

$$\mathrm{Let}\mathrm{me}\mathrm{try}\mathrm{sir}.\mathrm{Help}\mathrm{me}\mathrm{check}\mathrm{if}\mathrm{am}\mathrm{right}.$$

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