• Congratulations to the Class of 2024 on your results!
    Let us know how you went here
    Got a question about your uni preferences? Ask us here

James Ruse trig question (2 Viewers)

droungk

New Member
Joined
Apr 5, 2004
Messages
1
I am trawling through some old exams and this ruse question has taken advantage of me, i want my dignity back so can someone please help ?

A triangle ABC has angle C obtuse and and angle A twice that of angle B, all sides are integers.

Using the sine rule and the sin2@ and sin3@ formulae show that

a^2=b(b+c)

ps i dont know how to use computers they piss me of so much so i hope this works :chainsaw: <= (what the hell is that?)
 

nike33

Member
Joined
Feb 18, 2004
Messages
219
b/sin@ = a/2sin@cos@ = c/(3sin@-4sin^3@)
b = a/2cos@ = c/(4cos^2@-1)
a/b = 2cos@

b = c / ((a^2/b^2) - 1)
giving a^2 = b(b+c)

i wish we got qns like that in our papers
 

+Po1ntDeXt3r+

Active Member
Joined
Oct 10, 2003
Messages
3,527
Gender
Undisclosed
HSC
2003
ruse questions arent actualli beyond the syllabus.. cept for some Q8 and mofo questions in the yr 11 paper..

its the time management that will kill ppl.. and speed
the questions in ruse papers are usually straight forward..

just draw a diagram..

then write out sine rule.. sin 2@ n
then fill in the sine rule..

then use the sine 2@= 2sin@cos@
and sin 3@=3sin@-4sin^3@

it should be
b/sin@ = a/2sin@cos@ = c/(3sin(pi-@)-4sin^3(pi-@))

cos the pi-3@ is the angle size of C and the teachers in ruse usualli need the explanation that sin @ = sin(pi-@) since sine is cyclic every pi .. otherwise nike33 is right..
 
Last edited:

Users Who Are Viewing This Thread (Users: 0, Guests: 2)

Top