#Grade 10 Circles.
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Circle theorems will be of great help here, example:
PQRS is a cyclic quadrilateral
@gritty sundial
All the edges of the quadrilateral are touching the circle
Sum of opposite angles of a cyclic quadrilateral is 180°
so <PRQ will be 180-130?
Yup
okay so <POQ?
This
Angle subtended by a chord/arc at the centre is double the angle subtended by it at any other point on the circle
Oh so
2x50
or 2x <PRQ
how am i gonna find <OPQ tho
note that OP and OQ are radii of the circle
therefore OPQ is isoceles, with <OPQ = <OQP
we already know one angle
and what the angles in a trangle add up to
so just solve
no?
yes
The property of isosceles triangle and the property of cyclic quadrilateral and the angle at the centre theorem are very useful, so just keep them handy
Thanks for the reminder
ok
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Ping and say thanks
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@halcyon trail thanks
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another question idk if i can ask here
but
does 100 and 999 leave remainder 2 when divided by 7?
does not have to be the last division
find the least multiple less than that
if you want you can also use modulus properties to get the remainder
it has to be a 3 digit number
999÷7 = 142 with a reminder of 5
999-5 is the biggest 5 digit number divisible by 7
994
@gritty sundial
no like
999÷7
gives 2
wait
wdym gives 2
there's one 7 in 9
so
9-7
2
like that
U don't see the last digit only
U see the number as a whole
4 isn't divisible by 7 but 14 is
nono
the question isnt check if its divisible or not
can you send a screenshot of the problem
Yeah
I'll type it instead
That'll work
Consider the sequence of three digits numbers which leaves a remainder 2 on division by 7.
a) Find first term
b) Find last term
oh
arithmetic sequence
well 999 = 5 (mod 7)
so 2 (mod 7) = 5 - 3 ( mod 7)
or 999 - 3
so 996
996 = 2 (mod 7)
First term = 100
Last term= 996
102 leaves remainder of 4