Maths HELP as level trigonometry PLEASE NEED SERIOUS HELP

Hi I am stuck on a maths hwk and I dont understand all 3 parts of the question .
Here is the link to the question and btw it is question 4.
Thanks

Please edit your question and give us the link otherwise we can't help. Trigonometry is far too big a subject for us to give you any general help.

4 a)

cosine rule:
a2=b2+c2- 2bccosA

let's label each angle the letter at each apex of the triangle so angle theta is A and the length opposite angle A is length a.

Applying the cosine rule above:

6^2 = 5^2 + 4^2 - 2*5*4 cos A (which is theta)
36 = 25+16 - 40cos A
36 = 41-40cos A
36-41 = -40cosA
-5 = -40cos A
-5/-40 = cos A
1/8 = cos A

That's your first three marks.

b) If they ask me to show using a trig identity that something's going to have a root in it I instantly think of sin^2 theta + cos^2 theta = 1

We know from part a) that cos theata = 1/8

sin^2 theata + cos^2 theta = 1
sin^2 theta = 1 - cos^2 theta
sin^2 theta = 1 - (1/8)^2
sin^2 theta = 1 - 1/64
sin^2 theta = 63/64
sin theta = sqrt(63/64)
sin theta = sqrt(63)/sqrt(64)

now the sqrt(64) is easy, it's just 8

now sqrt(63) is harder, let's break it down

sqrt(63) = sqrt(9*7) = sqrt(9)*sqrt(7) = 3sqrt(7)

sin theta = 3sqrt(7)/8
Which is what part b says it should be so I'd say that is shown and worthy of the full 3 points.

The area of triangle ABC is really easy.

Area = 0.5 * 4 * 5 * sin theta

now we know that sin theta = 3sqrt(7)/8

4*5 = 20 and 0.5 * 20 = 10

So that's Area = 30 sqrt(7)/ 8

I hope that's worthy of both points.

Hope I helped.

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