Can you answer this question please

[Flamed-Ghoust]

http://papers.xtremepapers.com/CIE/Cambridge%20IGCSE/Mathematics%20(0580)/0580_s03_qp_2.pdf

Question 12 please

[NotAbod]

(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

[Flamed-Ghoust]

(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.

[NotAbod]

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

[Flamed-Ghoust]

Thank you so much,  +rep mate

[Flamed-Ghoust]

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.

[Flamed-Ghoust]

Come on i have my paper 4 today and i am not getting an answer yet?

[NotAbod]

Have you tried SOH CAH TOA?
ie sin (x) = opposite/hypotenuse etc

Sorry for the late reply, really busy with my university.

[Flamed-Ghoust]

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

[astarmathsandphysics]

See attachment

[NotAbod]

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

[Flamed-Ghoust]

how is angle ABX half od angle ABC?

[NotAbod]

It's a kite and that diagonal (DB) cut the smaller triangle (triangle ABC) into 2 equal halves therefore ABX x 2 = ABC

[Flamed-Ghoust]

oh, so u basically mean make an imaginery line ??

[NotAbod]

Yes and no.
Yes if it's not there.
No in this case because it's already there, line BD