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]

please 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!!

thx for it.

man, i still need help on those 2 questions :cry:

you heard of the double angle formulas?

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

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

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:

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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)

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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.