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# C3 Trigonometry question watch

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1. Hey, really stuck on this q, any help or clues to get started would be appreciated.

"In the diagram AB=6cm is the diamater of the circle and BT is the tangent to the circle at B. The chord AC is extened to meet this tangent at D and <DAB= theta
a) Show that CD = 6(sec theta - cos theta)"

If anyone has the edexcel C3 book the diagram is on p/82 and q.8
Thnaks
2. (Original post by jimber)
Hey, really stuck on this q, any help or clues to get started would be appreciated.

"In the diagram AB=6cm is the diamater of the circle and BT is the tangent to the circle at B. The chord AC is extened to meet this tangent at D and <DAB= theta
a) Show that CD = 6(sec theta - cos theta)"

If anyone has the edexcel C3 book the diagram is on p/82 and q.8
Thnaks
ok the first thing to do is connect C to B so that you have a triangle ACB where angle acb =90 since angle in a semicircle is 90.And triangle ABD is also 90 , diameter is perpendicular to tangent. Now using trig find AD for triangle ABC , which is 6sectheta , similary find AC in triangle ABD which is 6costheta.
so CD= 6(sectheta -costheta)
3. I got CD = 6* SinTheta*TanTheta by working out BD first

Presumably that is the same thing.
4. (Original post by rbnphlp)
ok the first thing to do is connect C to B so that you have a triangle ACB where angle acb =90 since angle in a semicircle is 90.And triangle ABD is also 90 , diameter is perpendicular to tangent. Now using trig find AD for triangle ABC , which is 6sectheta , similary find AC in triangle ABD which is 6costheta.
so CD= 6(sectheta -costheta)
Hey, thanks for the reply. Big help! I'll rep you in a mo
5. (Original post by jimber)
Hey, thanks for the reply. Big help! I'll rep you in a mo
thnx for the rep
6. (Original post by stevencarrwork)
I got CD = 6* SinTheta*TanTheta by working out BD first

Presumably that is the same thing.
Hey thanks, how did you work out BD? Just realised I have to use this method to answer the next part to the question lol
7. (Original post by jimber)
Hey thanks, how did you work out BD? Just realised I have to use this method to answer the next part to the question lol
DB and AB are at right angles. So BD/AB = tan Theta
8. (Original post by stevencarrwork)
DB and AB are at right angles. So BD/AB = tan Theta
Thanks, the next part of the question told me CD was 16cm and asked if i cud calculate AC, but i thought i would need the length of AD to do this?
9. You have a formula for CD, so presumably you solve it for theta.

What sort of triangle is ACB?
10. (Original post by stevencarrwork)
You have a formula for CD, so presumably you solve it for theta.

What sort of triangle is ACB?
I've got 6(sec theta - cos theta) = 16 but i'm not sure how to solve this, tried by multiplying by cos and replacing the cos's with x's and solving as a quadratic but it failed lol.

umm ACB is just a normal triangle, no right-angles but i get the feeling angle >CAB = >ABT?
11. If AB is a diameter, what angle is at C?
12. (Original post by stevencarrwork)
If AB is a diameter, what angle is at C?
DOH! i'm officialy retarded lol sorry about the late reply, but don't i still need to know one more thing before i can work out AC?
13. (Original post by jimber)
I've got 6(sec theta - cos theta) = 16 but i'm not sure how to solve this, tried by multiplying by cos and replacing the cos's with x's and solving as a quadratic but it failed lol.
It can't fail. The quadratic can be solved if you do what you did.
14. (Original post by stevencarrwork)
It can't fail. The quadratic can be solved if you do what you did.
x^2 - (16x/6) - 1 = 0
I tried to use the quadratic equation but it gave me answers >1
15. You must have a minus sign wrong.
16. (Original post by stevencarrwork)
You must have a minus sign wrong.
Yeah I did, thanks for your help, i'll rep you in a mo

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