A geometry problem by Mohammad Hamdar

Here, $6$ is equal to the diameter of the circle: $\dfrac{5}{\sin A} = 6 \Longrightarrow \sin A = \dfrac{5}{6}.$ By sine law, $\dfrac{AC}{\sin 60^{\circ}} = \dfrac{5}{\sin A} = 6 \Longrightarrow AC = 3\sqrt{3}.$ Now, $\begin{aligned} \angle C &= 180^{\circ} - \left(60^{\circ} + \sin^{-1} \left(\dfrac{5}{6}\right)\right) \\ &= 120^{\circ} - \sin^{-1} \left(\dfrac{5}{6}\right).\end{aligned}$ And, $\begin{aligned} \sin C &= \left(\dfrac{\sqrt{3}}{2}\right)\left(\cos\left(\sin^{-1} \left(\dfrac{5}{6}\right)\right)\right) - \left(-\dfrac{1}{2}\right)\left(\dfrac{5}{6}\right) \\ &= \left(\dfrac{\sqrt{3}}{2}\right)\left(\pm\dfrac{\sqrt{11}}{6}\right) - \left(-\dfrac{1}{2}\right)\left(\dfrac{5}{6}\right) \\ &= \dfrac{\pm\sqrt{33} + 5}{12}. \end{aligned}$ So we take the positive one. Thus, the area of $\triangle ABC$ is $\dfrac{1}{2}\left(3\sqrt{3}\right)\left(5\right)\left(\dfrac{\sqrt{33} + 5}{12}\right) = \dfrac{5\sqrt{3}\left(5 + \sqrt{33}\right)}{8}.$ Finally, we get $a + b + c + d$ which is $\boxed{49}$ .

Note:

$\begin{aligned} \cos\left(\sin^{-1}\left(\dfrac{5}{6}\right)\right) &= \pm\sqrt{1 - \sin^2\left(\sin^{-1}\left(\dfrac{5}{6}\right)\right)} \\ &= \pm\sqrt{1 - \left(\dfrac{5}{6}\right)^2} \\ &= \pm\dfrac{\sqrt{11}}{6} \end{aligned}$

$\text{Let O be the center of the circle, and} \ \angle\ OCB=X.\\ \therefore\ \color{#3D99F6}{ CosX=\dfrac{\frac 5 2} 3=\dfrac 5 6,\ and \ SinX=\dfrac{\sqrt{11}} 6. }\\ Since\ \angle \text{ ABC=60, tangent at C makes 60 with AC ,} \\ \text{so angle OCA= 90 - 60 = 30 and angle BCA=X+30. }\\ AC=2 *( 3 * Cos30)=\dfrac{6\sqrt3} 2=3\sqrt3.\\ Area\ \Delta\ ABC=\frac 1 2*AC*BC*SinBCA=\frac 1 2*5*3\sqrt3*SinBCA,\ \ \ \ \ Expanding\ \ Sin,\\ Area\ \Delta\ ABC=\frac 1 2*5*3\sqrt3* \left( \dfrac{\sqrt{11}} 6*\dfrac{\sqrt3} 2+\dfrac 5 6*\dfrac 1 2 \right )\\ Area\ \Delta\ ABC=\dfrac{5\sqrt3(\sqrt{33}+5)} 8 \ \\ \therefore a+b+c+d= 5+3+33+8=49.$

Write $R=abc/4\triangle$ , where $\triangle$ represent area of Triangle.

$a=5$ .............. $given$

Using this we get area of triangle , $\triangle = 5bc/12$ , under usual notations. ......... $eq^{n} 1$

Since $b=2R sinB = 3\sqrt{3}$ , by extended sine rule.

Now using Cosine rule for angle $B$ , and putting values of a & b we get

$c$ = $5+\sqrt{33}/2$

Now putting it in equation 1, We get

$\triangle= 5\sqrt{3}(5+\sqrt{33})/8$ .

Hence by comparing we get

$a+b+c+d= \boxed{\boxed{49}}$

Note : Sine Rule (extended) $a/sinA = b/sinB = c/sinC = 2R$

Cosine rule :

$cosA= b^{2} + c^{2} -a^{2} /2bc$