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rinoa1989
Group Title
Hi I need help for a question here related to Circle as attached below. Please help Thanks! ><"
 one year ago
 one year ago
rinoa1989 Group Title
Hi I need help for a question here related to Circle as attached below. Please help Thanks! ><"
 one year ago
 one year ago

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rinoa1989 Group TitleBest ResponseYou've already chosen the best response.0
The answer given is A
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
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 ?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
i meant length of arc q and angle p...
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
you can create isocolese triangles from central angles.dw:1359728013730:dw
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
for your question, i extended WO to meet circle at say L dw:1359727980102:dw
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
i think working the iso triangle with the 65 would be simpler :)
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
dw:1359728146814:dw
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
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...
 one year ago

rinoa1989 Group TitleBest ResponseYou've already chosen the best response.0
omg looks so complicated! LOL ><"
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
yeah...i figured arc WV = 50 using similar thing, which is essentially similar to what u did..
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
my way... arc WV+VU+UL = 180 but since angle LWV = 65 , VU+UL = 130 which gives WV = 50 then 2x = 70 +WV
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
ask if any doubts...
 one year ago

rinoa1989 Group TitleBest ResponseYou've already chosen the best response.0
wait let me digest first lol
 one year ago

rinoa1989 Group TitleBest ResponseYou've already chosen the best response.0
i duno how u get this arc WV+VU+UL = 180
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
since is O the centre, arc WL is semicircle WL = 180 WV+VU+UL = 180
 one year ago

rinoa1989 Group TitleBest ResponseYou've already chosen the best response.0
ic ok how bout this one 2x = 70 +WV ??
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
dw:1359729747346:dw so 2x = 70+WV
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
thats why i first asked you whether you know that property....
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
still don't get it ?
 one year ago

rinoa1989 Group TitleBest ResponseYou've already chosen the best response.0
sorry i duno about that property i tot it is related to the photo i attach here
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
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....
 one year ago

rinoa1989 Group TitleBest ResponseYou've already chosen the best response.0
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! ><"
 one year ago
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