1. ## Trigonometry

There is my problem in attachment. I've done (a), but how can I solve (b). Please help me.

2. Originally Posted by geton
There is my problem in attachment. I've done (a), but how can I solve (b). Please help me.
Let the point E be the centre of the circle and consider triangle AEC.
AE = EC = 3 cm.
Solve $16 = 6(\sec \theta - \cos \theta)$ for $\cos \theta$ and hence $\theta$.
Since triangle AEC is isosceles, the angle at E is equal to $(180 - 2 \theta)^o$.
Now solve for AC using the cosine rule.

3. Originally Posted by mr fantastic
Let the point E be the centre of the circle and consider triangle AEC.
AE = EC = 3 cm.
Solve $16 = 6(\sec \theta - \cos \theta)$ for $\cos \theta$ and hence $\theta$.
Since triangle AEC is isosceles, the angle at E is equal to $(180 - 2 \theta)^o$.
Now solve for AC using the cosine rule.
Note: Since you can get an exact value for $\cos \theta$, it's possible to get an exact value for $\cos(180 - 2 \theta)$.
Do so if you're required to get an exact value for AC ......
Otherwise use a calculator to get a decimal approximation.

4. Hello, geton!

Draw chord $BC.$

Since $\Delta ACB$ is inscribed in a semicircle, $\angle ACB = 90^o$

In right triangle $DBA\!:\;\;\sec\theta \:=\:\frac{AD}{6}\quad\Rightarrow\quad AD \:=\:6\sec\theta$

In right triangle $ACB\!:\;\;\cos\theta \:=\:\frac{AC}{6}\quad\Rightarrow\quad AC \:=\:6\cos\theta$

Hence: . $CD \;=\;AD - AC \;=\;6\sec\theta - 6\cos\theta$

. . Therefore: . $(a)\;\;CD\;=\;6(\sec\theta - \cos\theta)$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

In right triangle $ACB$, we have:
. . $\cos\theta = \frac{AC}{6}\quad\Rightarrow\quad AC \:=\:6\cos\theta$ .[1]

Since $CD = 16$, part (a) gives us: . $6(\sec\theta - \cos\theta) \:=\:16$

Then we have: . $3\left(\frac{1}{\cos\theta} - \cos\theta\right) \:=\:8$

. . which simplifies to: . $3\cos^2\!\theta + 8\cos\theta - 3 \:=\:0$

. . which factors: . $(3\cos\theta-1)(\cos\theta +3) \:=\:0$

And we have: . $3\cos\theta - 1 \:=\:0\quad\Rightarrow\quad \cos\theta \:=\:\frac{1}{3}$ .**

Substitute into [1]: . $AC \:=\:6\left(\frac{1}{3}\right)\quad\Rightarrow\qua d\boxed{(b)\;AC \:=\:2}$

**
The other factor produces: . $\cos\theta+3 \:=\:0\quad\Rightarrow\quad\cos\theta \:=\:-3$
. . which has no real root.