# Thread: circle and triangle trig help

1. ## circle and triangle trig help

In the diagram AB=6 cm 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=θ.
(a) Show that CD=6(secθ−cosθ) .
(b) Given that CD=16 cm, calculate the length of the chord AC.

2. Originally Posted by Tweety
In the diagram AB=6 cm 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=θ.
(a) Show that CD=6(secθ−cosθ) .
(b) Given that CD=16 cm, calculate the length of the chord AC.

(a) first , we find AD .

$\displaystyle cos\theta=\frac{6}{AD}$

$\displaystyle AD=6sec\theta$

Then now AC .

$\displaystyle sin(90-\theta)=\frac{AC}{6}$

$\displaystyle AC=6cos\theta$

$\displaystyle AD-AC=CD$

$\displaystyle CD=6sec\theta-6cos\theta$

$\displaystyle CD=6(sec\theta-cos\theta)$

3. Originally Posted by Tweety
In the diagram AB=6 cm 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=θ.
(a) Show that CD=6(secθ−cosθ) .
(b) Given that CD=16 cm, calculate the length of the chord AC.
(a) The key to this is the 'angle in a semicircle is a right-angle' property; i.e. $\displaystyle \angle ACB = 90^o$. And, of course, the angle between a radius (or a diameter) and a tangent is also a right-angle. So $\displaystyle \angle ABD = 90^o$

So in $\displaystyle \triangle ABC, AC = 6 \cos \theta$

and in $\displaystyle \triangle ABD, AD = 6\sec\theta$

$\displaystyle \Rightarrow CD = 6(\sec\theta - \cos\theta)$

For part (b): $\displaystyle 16 = 6(\sec\theta - \cos\theta)$

$\displaystyle \Rightarrow 3\sec\theta - 3\cos\theta = 8$

$\displaystyle \Rightarrow 3 - 3\cos^2\theta = 8\cos\theta$

$\displaystyle \Rightarrow 3\cos^2\theta + 8\cos\theta - 3 = 0$

$\displaystyle \Rightarrow (3\cos\theta - 1)(\cos\theta + 3)=0$

$\displaystyle \Rightarrow \cos\theta = \tfrac13$ ($\displaystyle -3$ being impossible)

$\displaystyle \Rightarrow AC =6\cos\theta = 2$

(a) first , we find AD .

$\displaystyle cos\theta=\frac{6}{AD}$

$\displaystyle AD=6sec\theta$

Then now AC .

$\displaystyle sin(90-\theta)=\frac{AC}{6}$

$\displaystyle AC=6cos\theta$

$\displaystyle AD-AC=CD$

$\displaystyle CD=6sec\theta-6cos\theta$

$\displaystyle CD=6(sec\theta-cos\theta)$
thanks,
but I can't seem to follow from this step
$\displaystyle sin(90-\theta)=\frac{AC}{6}$

how did you get this?

5. Originally Posted by Tweety
thanks,
but I can't seem to follow from this step
$\displaystyle sin(90-\theta)=\frac{AC}{6}$

how did you get this?
I assume that it's from the right triangle ABC. Angle ABC would be $\displaystyle 90^{\circ} - \theta$.

01

6. Originally Posted by Tweety
thanks,
but I can't seem to follow from this step
$\displaystyle sin(90-\theta)=\frac{AC}{6}$

how did you get this?

erm ok , from your diagram , you draw another line from C to B and angle ACB is 90 degrees so angle ABC is 90-theta
Then you will get :

$\displaystyle sin(90-\theta)=\frac{AC}{6}$

erm ok , from your diagram , you draw another line from C to B and angle ACB is 90 degrees so angle ABC is 90-theta
Then you will get :

$\displaystyle sin(90-\theta)=\frac{AC}{6}$
why is it $\displaystyle sin\theta$ ? should it not be $\displaystyle cos\theta$?

thats what i dont get;

Thanks.

8. Originally Posted by Tweety
why is it $\displaystyle sin\theta$ ? should it not be $\displaystyle cos\theta$?

thats what i dont get;

Thanks.

$\displaystyle sin(90-\theta)=cos\theta$