Dr. Fiendish 3

$\cos \angle {OBC}=0.9\implies \angle {OBC}\approx 25.842\degree\implies \angle {BOC}\approx 128.316\degree\implies \angle {CAB}\approx 64.158\degree$

$\cos \angle {OBA}=0.7\implies \angle {OBA}\approx 45.576\degree\implies \angle {AOB}\approx 88.849\degree\implies \angle {ACB}\approx 44.424\degree\implies \angle {ABC}\approx 71.418\degree$

$|\overline {CD}|=9\sin \angle {ABC}\approx 8.531$

Hence the required answer is $64.158+8.531=\boxed {72.689}$ .

Since $\angle ACB$ is at the circumference and $\angle AOB$ at the center extended from the same chord $AB$ , $\angle ACB = \dfrac 12 \angle AOB$ and $\sin \angle ACB = \dfrac {3.5}5 = 0.7$ . By sine rule , we have:

$\begin{aligned} \frac {\angle CAB}9 & = \frac {\angle ACB}7 \\ \sin \angle CAB & = 9 \cdot \frac {0.7}7 = 0.9 \\ \implies \angle CAB & \approx 1.119769515 \text{ rad} \approx 64.158^\circ \end{aligned}$

Then we have

$\begin{aligned} CD & = BC \sin \angle ABC \\ & = 9 \sin (\pi - \angle ACB - \angle CAB) \\ & = 9 \sin (\angle ACB + \angle CAB) \\ & = 9 \sin (\sin^{-1} 0.7 + \sin^{-1} 0.9) \\ & = 9 (0.7\sqrt{1-0.81} + 0.9\sqrt{1-0.49}) \\ & = 9 (0.7\sqrt{0.19} + 0.9\sqrt{0.51}) \\ & \approx 8.531 \end{aligned}$

Therefore the answer is $64.158+8.531 \approx \boxed{72.7}$ .

Note that we have to check that whether $O$ is in $\triangle ABC$ , as the question requires so.
Apply the cosine rule on $\triangle ABO.$ $7^2=5^2+5^2(radii,OA=OB)-2\times5\times5\cos \angle BOA.$ $\Rightarrow \cos \angle BOA=0.02, \angle BOA=\cos^{-1} 0.02=88.854008…^\circ$ and since angle at center is twice angle on circumference , $\angle BCA=\frac{\angle BOA}{2}=44.427004…^\circ$
Now apply the sine rule on $\triangle ABC$ . $\frac{\sin \angle BCA=0.7}{AB=7}=\frac{\sin \angle CAB}{BC=9}$ $\Rightarrow \angle CAB=\sin^{-1} 0.9=64.158067236833…^\circ$ Now let’s find $CD$ . Note that $CD$ is the height on $AB$ . Hence we have to find the area of $\triangle ABC$ , by using the sine rule or the cosine rule and applying $S_{triangle}=ab\sin C=\frac{abc}{4R}=\frac{r(a+b+c)}{2}$ or Heron’s formula . I would suggest using $S=ab \sin C$ because it only requires the angle ABC.
So the area of $\triangle ABC$ = $59.71464353085…$
$\Rightarrow CD=\frac{59.71464353085…}{7}=8.53066336155…$
Check! $O$ is in $\triangle ABC$ .
So the answer is $64.158067236833…+8.53066336155…=\boxed{72.688730598383…}.$

What I did was applied the extended law of sines to get the angles of the triangle and use that to find the length of CD

Overview of a (Bashy) Solution: We are given two sides of the triangle, and can figure out the third side from the formula $\frac{abc}{4R}$ $= area$ . From there, use Pythagorean Theorem to find $CD$ , and trigonometry to find $\angle CAB$ .

Step 1: Find $AC$ .

Let $AC = x$ . Then we have

$\frac{abx}{4R}$ $= area$

Using Heron's formula for the right side gives

$\frac{x⋅7⋅9}{4⋅5}$ $=$ $\frac{1}{2^2}$ $⋅$ $\sqrt{(16+x)(x-2)(x+2)(16-x)}$

Solve to get $x≈9.4785...$ or $x≈3.37605...$

Step 2:Pythagorean Theorem on $\Delta ADC$ and $\Delta CDB$ to find $CD$

Let $AD=w.$

Then we have the two equations

$x^2-w^2=CD^2$

$9^2-(7-w)^2=CD^2$

Solving we have $w=$ $\frac{x^2-32}{14}$ , so the only choice of $x$ that makes $w$ positive is $x≈ 9.4785$ .

Plugging in,

$w≈4.13158$ so

$CD=$ $\sqrt{x^2-w^2}≈8.53066$

Step 3: Find $\angle CAB$ .

$\tan \angle CAB$ = $\frac{CD}{w}$ = $\frac{8.53066}{4.13158}$

$\angle CAB$ = $64.15806 ^\circ$

Summing, $8.53066+64.15806≈\boxed{72.6887}$