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secondary 3 | A Maths
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PLS HELP ASAP CANT EVEN SOLVE(i) N HV BEEN STARING AT THIS FOR TWO DAYS
Draw the radius OU first.
SVU = 180° - STU (opposite angles of a cyclic quadrilateral add up to 180°. SVUT is the cyclic quadrilateral in this case)
OVU = SVU (common angle)
OV = OU (both are radius of same circle)
So triangle OVU is isosceles.
OVU = OUV (base angles of isosceles triangle)
SOU = OVU + OUV (exterior angle = sum of 2 interior opposite angles)
= 2 OVU (since OUV = OVU)
b)
OUP and OSP = 90°
(tangent perpendicular to radius)
SPU = 360° - OUP - OSP
(sum of angles in a quadrilateral add up to 360°, the quadrilateral in this case is SOUP)
c)
TSP = TWS
(angles in alternate segments
VST = VSP - TSP (adjacent angles of the same angle VSP)
VSP = OSP = 90° (common angle)
so VST = OSP - TSP
d)
VUT = 180° - VST
(Opposite angles of a cyclic quadrilateral add up to 180°. SVUT is the cyclic quadrilateral in this case)
As Eric has pointed out, its now:
Angles in opposite segments are supplementary
It’s currently phrased as
“angles in opposite angles are supplementary”.
For quite a while already I was wondering why this phrase has not been used before, considering that “angles in the same segment are equal” already suggests that drawing a fourth point on the other side of the chord instead would make the angles automatically in opposite segments.
J and Eric: Honestly cannot say with certainty whether older, factually correct reasons will be accepted by Cambridge examiners, but why take the risk? I advise my students to follow the phrasing used within the syllabus, whether math or science.
The latest syllabus is available for download as a PDF from www.seab.gov.sg
SEAB = Singapore Examinations and Assessment Board, which oversees all National Exams from PSLE to the A-Levels.
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