Please help me with part (ii) of the question attached below :) I've included my working out also: part 1 -http://prntscr.com/844fkr part 2 - http://prntscr.com/844g0b

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Please help me with part (ii) of the question attached below :) I've included my working out also: part 1 -http://prntscr.com/844fkr part 2 - http://prntscr.com/844g0b

Mathematics
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to conclude, you may use this fact of isosceles triangle : perpendicular bisector of a side passes through the opposite vertex
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show that \(\angle B = 60^{\circ}\)

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ahhhhhhhh okay! It was so simple x.x \[\tan \theta = \frac{ AM }{ BM }\] \[= ( R + \frac{ R }{ 2 }) \times \frac{ 1 }{ R + \sqrt{R^2 - \frac{ R^2 }{4}} }\] \[= \frac{ 3R }{ 2} \times \frac{ 1 }{ \frac{ R \sqrt{3} }{ 2 } }\] \[= \frac{ 3R }{ \sqrt{3}} = \sqrt{3}\] \[\therefore \theta = 60 degrees \]
thaaaanks ganeshie! ^_^
np

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