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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
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

This Question is Closed

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1to conclude, you may use this fact of isosceles triangle : perpendicular bisector of a side passes through the opposite vertex

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1dw:1439465636872:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1show that \(\angle B = 60^{\circ}\)

butterflydreamer
 one year ago
Best ResponseYou've already chosen the best response.1ahhhhhhhh okay! It was so simple x.x \[\tan \theta = \frac{ AM }{ BM }\] \[= ( R + \frac{ R }{ 2 }) \times \frac{ 1 }{ R + \sqrt{R^2  \frac{ R^2 }{4}} }\] \[= \frac{ 3R }{ 2} \times \frac{ 1 }{ \frac{ R \sqrt{3} }{ 2 } }\] \[= \frac{ 3R }{ \sqrt{3}} = \sqrt{3}\] \[\therefore \theta = 60 degrees \]

butterflydreamer
 one year ago
Best ResponseYou've already chosen the best response.1thaaaanks ganeshie! ^_^
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