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Solution for 2008 hsc q10 (1 Viewer)
- Thread starter chopstick
- Start date
random-1005
Banned
- Joined
- Dec 15, 2008
- Messages
- 609
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- Male
- HSC
- 2009
its easy, its just similar triangles, ratio of sides and A=1/2absinC
Fluorescent
Banned
Q10) Bi
ANGLE OMP = ANGLE OJK
ANGLE POM = ANGLE OJK (ALTERNATE ANGLES MP||KJ)
ANGLE MOP = ANGLE JOK (VERT OPP)
THEREFORE Triagnle MOP||| Triangle JOK (equiangular)
therefore MP/JK = OM/OJ (corresponding sides of similar triangles in the same ratio)
therfore MP/S = (l-x)/x
mp= s(l-x)/x
therefore A= area triangle okj+area triangle omp
= (1/2ln) x s x (X) x sin alpha + 1/2 x (s(l-x)/x) x (l-x) x sin alpha
= 1/2 sx sin alpha + (1/2x) x (s(l-x)^2) sin alpha
= s sin alpha (x/2 + ( l^2 - 2lx+x^2)/2x)
= s (x^2+ l^2 - 2lx+x^2)/2x) sin alpha
= s (l^2/2x - l +x )sin alpha
= s (x - l +l^2/2x) sin alpha
done...
holy hell i hope i didnt make mistakes
ANGLE OMP = ANGLE OJK
ANGLE POM = ANGLE OJK (ALTERNATE ANGLES MP||KJ)
ANGLE MOP = ANGLE JOK (VERT OPP)
THEREFORE Triagnle MOP||| Triangle JOK (equiangular)
therefore MP/JK = OM/OJ (corresponding sides of similar triangles in the same ratio)
therfore MP/S = (l-x)/x
mp= s(l-x)/x
therefore A= area triangle okj+area triangle omp
= (1/2ln) x s x (X) x sin alpha + 1/2 x (s(l-x)/x) x (l-x) x sin alpha
= 1/2 sx sin alpha + (1/2x) x (s(l-x)^2) sin alpha
= s sin alpha (x/2 + ( l^2 - 2lx+x^2)/2x)
= s (x^2+ l^2 - 2lx+x^2)/2x) sin alpha
= s (l^2/2x - l +x )sin alpha
= s (x - l +l^2/2x) sin alpha
done...
holy hell i hope i didnt make mistakes
Aquawhite
Retiring
I have a printed solution to it from somewhere but my scanner/printer is currently "OUT OF ORDER" ... sorry. ^ Post above looks correct.