IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => IGCSE/ GCSE => Math => Topic started by: Flamed-Ghoust on November 02, 2013, 09:22:00 am
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http://papers.xtremepapers.com/CIE/Cambridge%20IGCSE/Mathematics%20(0580)/0580_s03_qp_2.pdf
Question 12 please
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(a)
angle ABC is 90 degrees.
90 + 2p + 3p = 180
p = (180-90)/5
p = 18
(b)
AEDC is a cyclic quadrilateral, therefore opposite angles = 180
Angle CAE + CDE = 180
q + 5q = 180
q = 180/6
q = 30
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(a)
angle ABC is 90 degrees.
90 + 2p + 3p = 180
p = (180-90)/5
p = 18
Thank you very much NotAbod, but how do you know the third angle for part "a" is 90 degrees, it doesn't say so in the question. And a square is not drawn in the diagram that will prove that it is equal to 90 degrees.
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You must learn circle theorems.
ABC is a cyclic triangle. AOC is Diameter.
Let Angle ABC = x
then AOC is 2x.
AOC is diameter and its angle is 180
ABC x 2 = AOC
2x = 180
x = 180/2
x = 90
Take a look at this and learn them: http://www.mathsisfun.com/geometry/circle-theorems.html
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Thank you so much, ^-^ +rep mate
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Can you help me now with this question please.
I found the area but could not find the angle, so in the exam i wrote it 60 degrees even though i doubt it is. ;D :-*
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Come on i have my paper 4 today and i am not getting an answer yet????
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Have you tried SOH CAH TOA?
ie sin (x) = opposite/hypotenuse etc
Sorry for the late reply, really busy with my university.
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the triangle is not 90 degrees, so the SOH CAH TOA wont work... the sin rule also wont work cause i only have 1 angle and that angle is unknown, the cosine rule also wont work cause we only have one side and the other 2 are unknown, the biggest problem is that the intersection of AC isnt equal in size
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See attachment
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the triangle is not 90 degrees, so the SOH CAH TOA wont work... the sin rule also wont work cause i only have 1 angle and that angle is unknown, the cosine rule also wont work cause we only have one side and the other 2 are unknown, the biggest problem is that the intersection of AC isnt equal in size
AXB is 90 Degrees and I assume you know XB and AC is given as 12cm
Use SOHCAHTOA to find angle ABX (which is 1/2 of ABC)
ABX x 2 = ABC
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how is angle ABX half od angle ABC????
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It's a kite and that diagonal (DB) cut the smaller triangle (triangle ABC) into 2 equal halves therefore ABX x 2 = ABC
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oh, so u basically mean make an imaginery line ??
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Yes and no.
Yes if it's not there.
No in this case because it's already there, line BD
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Too easy a question to take so long.
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Yes and no.
Yes if it's not there.
No in this case because it's already there, line BD
But line BD isn't dividing the triangle into 2 equal parts ??? so how??
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Yes it does, because it is a KITE.
The diagram is not drawn to scale
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So in a kite the line BD always divides it into 2??
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BD is the line of symmetry and the top half of the kite is the same as the bottom half of the kite.
If it's a kite, then yes, that line divides the upper and bottom halves into 2 equal parts.
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i should have known, in the exam i just wrote 60 degrees. guess i am wrong
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60 is a good guss. Better than not guessing at all.
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I only had 3 minutes so I well I guess 60 is a good guess to bad it was a wrong one
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It was a good guess. When you guess aim for the middle.
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hope I get a mark for guessing. my teacher said they would probably cut marks for me messing up the whole page with calculations and then crossing them
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You wont get a mark for guessing if you guess wrong.
You get one mark for a coreect answer and marks for any method