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Jul 4th, 2006, 09:24 AM
#1
Thread Starter
New Member
Trigonometry [please i need help!]
Trigonometry
Q1 : Given that tan 2y = -0.7265, find all the values of y between 0 degree and 360 degree.
Q2: attached picture. [P1010892]
Solution of Triangles
Q1: attached picture. [P1010893]
Last edited by tecm0; Jul 10th, 2006 at 01:31 AM.
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Jul 6th, 2006, 07:28 AM
#2
Thread Starter
New Member
Re: New to Trigonometry / Solution of Triangles. [HELP!]
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Jul 6th, 2006, 12:20 PM
#3
Re: New to Trigonometry / Solution of Triangles. [HELP!]
for trig, Q2, (i am using x instead of theta)
we know that: sin2x + cos2x = 1
putting value of sinx: 25/169 + cos2x = 1
therefore: cos2x = 1 - 25/169
..............cos2x = 144/169
..............cosx = 12/13
and: secx = 1/cosx
so: secx = 13/12
and: cotx = cosx/sinx
so: cotx = 12/13 * -13/5 = -12/5
the first trig question is a bit confusing, atleast to me!!
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Jul 10th, 2006, 01:32 AM
#4
Thread Starter
New Member
Re: Trigonometry [please i need help!]
thx for it.
man, i still need help on those 2 questions
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Jul 10th, 2006, 08:14 AM
#5
Addicted Member
Re: Trigonometry [please i need help!]
you heard of the double angle formulas?
Rich
A)bort, R)etry, I)nfluence with large hammer.
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Jul 10th, 2006, 11:59 AM
#6
Member
Re: Trigonometry [please i need help!]
2y = arctan(-0.7265) = -35.99840...
But the tangent function is periodic, so arctan(-0.7265) also equals 144.00159..., 324.00159..., 504.00159..., 684.00159... .
But this equals 2y, so halve all the values to get y:
y = 72.0, 162, 252 and 342 degrees to 3 s.f. in the range 0<y<360
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Jul 10th, 2006, 12:04 PM
#7
Member
Re: Trigonometry [please i need help!]
For your other question, use the cosine rule 3 times and simply state the largest angle you find. If the sides a,b and c are opposite angles A, B and C:
a² = b² + c² - 2bc.cosA
b² = a² + c² - 2ac.cosB
c² = a² + b² - 2ab.cosC
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Jul 10th, 2006, 12:08 PM
#8
Re: Trigonometry [please i need help!]
the second question, finding the angle.
well there are 2 methods, first one is using the Laws of Sines which states, that the length of each side is proportional to the SINE of the angle opposite it. but i am unable to calculate from it. maybe i am doing something wrong.
another method would be, use Hero's formula to find the area of the triangle, and then...
let me give you an example
Code:
Hero's Formula
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K = sqrt(s*[s-a]*[s-b]*[s-c]), where s = (a+b+c)/2
so according to your question, s = 11/2 or 5.5 and K = 15.2 or 3.8
also Area of the triangle is also given by:
K = (1/2)*a*b*sin(C),
therefore, sin(C) = 2*K/(a*b)
Similarly
sin(A) = 2*K/(b*c),
sin(B) = 2*K/(c*a)
hope it helps you.
EDIT: here a,b,c are the sides and A,B,C are the angles opposite to a,b,c respectively.
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