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Applying C3 Trigonometry to triangles

Hi all, I am going through C3 trigonometry and it going fairly well. However, when I get a question that applies all the stuff in C3 to an actual triangle I am lost. There seems to be only a handful of these type of questions in the Edexcel Student Book. There is much more focus on learning all the indenties: addition, double angle etc; and solving equations.

Question 8, Eexercise 6c:

In the diagram AB = 6cm is the diameter of the circle and BT is the tangent to the circle at B. The chord AC is extended to meet this tangent at D and DAB = x.

a) show that CD = 6(secx - cosx)

b)Given that CD = 16cm, calculated the length of the chord AC

I have attached the picture. Sorry, it horizontal.

Also, question 10) Exercise 7b

In ABC, AB = 4cm, AC = 5cm, ABC = 2x amd ACB = x, Find the value of x?

Another question using application to a triangle and I am lost.I wish Examsolutions or Hegartymaths gave an example, but they don't.

Thank you so much for all your help.

IMAG0376.jpg166.0KB

Reply 1

Tygra

Original post by Homletmoo

In the first part of the question, find the length AD, then subtract the length AC (hint: consider the triangle ACB: you can use alternate segment theorem to find its internal angles).

In the second part, set 16 = 6(secx - cosx), and (noticing that this is a hidden quadratic in secx) solve for secx. You should have found the right equations at this point to just plug this in somewhere and get AC.

Hopefully that's enough to get you started. :smile:

Thank you bro, with your help I just whipped that question's arse. The chord AC = 2cm! Very pleased about that, thanks.

I tried the question. I used the sine rule. but got the wrong answer.

This was how I set it up:

sinx/4 = 1-2sin^2x/5

I used the quadratic formula and got 27.4 degrees. This is wrong.

Is it possible to help me out with the second question I posted about?

(edited 8 years ago)

Reply 2

Tygra

Original post by Homletmoo

Use sine rule to get an equation in sinx and sin2x, use double angle formula on the sin2x, factorise and solve for x. There should be two solutions, but one of them is pretty trivial.

Homletmoo, that is what I have done isn't it?

Reply 3

Tygra

Original post by Homletmoo

Sorry, my reading comprehension is awful, I completely missed your working and only read the last sentence...

You appear to have used the double angle formula for cos2x rather than sin2x.

That is okay. Yes, when I went to bed last night I suddenly twigged that I used the double angle formula for cos2x, not sin2x. I have the answer now. Thank you again.

Do you have any advice on complacency? As you have seen a stupid silly mistake was my downfall. I think complacency is my biggest enemy in maths. :frown: