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rinoa1989
Hi I need help for a question here related to Circle as attached below. Please help Thanks! ><"
The answer given is A
do you know the central angle and corresponding arc property ...like |dw:1359727764547:dw| if central angle given is 70 degrees ,say, then can you find length of arc p and angle q ?
i meant length of arc q and angle p...
you can create isocolese triangles from central angles.|dw:1359728013730:dw|
for your question, i extended WO to meet circle at say L |dw:1359727980102:dw|
i think working the iso triangle with the 65 would be simpler :)
|dw:1359728146814:dw|
oh, but i figured it out the using other method, i think..and i donno how that would help...maybe you can proceed to show us...
omg looks so complicated! LOL ><"
the base angles of an iso tri are equal; leaving the central angle to be 180 - 2base; in this case, 180 - 2(65) total arc length is then 70 + 180 - 2base and x is determined to be half that
yeah...i figured arc WV = 50 using similar thing, which is essentially similar to what u did..
my way... arc WV+VU+UL = 180 but since angle LWV = 65 , VU+UL = 130 which gives WV = 50 then 2x = 70 +WV
wait let me digest first lol
i duno how u get this arc WV+VU+UL = 180
since is O the centre, arc WL is semicircle WL = 180 WV+VU+UL = 180
ic ok how bout this one 2x = 70 +WV ??
|dw:1359729747346:dw| so 2x = 70+WV
thats why i first asked you whether you know that property....
sorry i duno about that property i tot it is related to the photo i attach here
its the same property i illustrated.. angle A = x= 1/2 of central angle O(2x) = 1/2 of arc QP so, here x = half of arc TV x= TV/2 2x=TV = 70+WV....
oops ic ok! coz i just use the 70 to divide 2 to find x now i understand thank you so much for your time in explaining! ><"