## butterflydreamer one year ago 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

1. ganeshie8

to conclude, you may use this fact of isosceles triangle : perpendicular bisector of a side passes through the opposite vertex

2. ganeshie8

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3. ganeshie8

show that $$\angle B = 60^{\circ}$$

4. butterflydreamer

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

5. butterflydreamer

thaaaanks ganeshie! ^_^

6. ganeshie8

np