# circle and triangle trig help

• Jul 2nd 2009, 07:34 AM
Tweety
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.
• Jul 2nd 2009, 07:50 AM
Quote:

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 .

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

$AD=6sec\theta$

Then now AC .

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

$AC=6cos\theta$

$AD-AC=CD$

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

$
CD=6(sec\theta-cos\theta)
$
• Jul 2nd 2009, 07:54 AM
Quote:

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. $\angle ACB = 90^o$. And, of course, the angle between a radius (or a diameter) and a tangent is also a right-angle. So $\angle ABD = 90^o$

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

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

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

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

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

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

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

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

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

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

• Jul 2nd 2009, 08:05 AM
Tweety
Quote:

(a) first , we find AD .

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

$AD=6sec\theta$

Then now AC .

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

$AC=6cos\theta$

$AD-AC=CD$

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

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

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

how did you get this?
• Jul 2nd 2009, 08:10 AM
yeongil
Quote:

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

how did you get this?

I assume that it's from the right triangle ABC. Angle ABC would be $90^{\circ} - \theta$.

01
• Jul 2nd 2009, 08:14 AM
Quote:

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

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 :

$sin(90-\theta)=\frac{AC}{6}$
• Jul 2nd 2009, 08:20 AM
Tweety
Quote:

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 :

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

why is it $sin\theta$ ? should it not be $cos\theta$?

thats what i dont get;

Thanks.
• Jul 2nd 2009, 08:26 AM
why is it $sin\theta$ ? should it not be $cos\theta$?
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