A nice result

Use this . You will find that $\overline{AD} = \dfrac{6\sqrt{2}(\sqrt{3}+1)}{4\sqrt{2}+3}$ .

By cosine law on $\triangle ABC$ , we have

$BC^2=3^2+4^2-2(3)(4)(\cos 75)$

$BC\approx 4.334552216$

By sine law on $\triangle ABC$ , we have

$\dfrac{\sin B}{4}=\dfrac{\sin 75}{4.334552216}$

$B\approx 63.04630206^\circ$

Thus, $\angle ADB=180-30-63.04630206\approx 86.95369794^\circ$

By sine law on $\triangle ABD$ , we have

$\dfrac{AD}{\sin 63.04630206}=\dfrac{3}{\sin 86.95369794}$

$AD\approx 2.67790$

Area Tr. ABC = Area Tr. BAD + Area Tr, CAD. Hence, (1/2) 3 4 sin(75)=(1/2)[3 x sin(30)+4 x*sin(45)] or x= 6(√2+√6)/(3+4√2) or x ~= 2.6779 cm

(1) Compute BC by Law of Cosines. (2) Compute total area by Herron's formula. (3) Compute total area by A = (1/2)(AD)[((AB)sin(30) + (AC)sin(45)). Equate the expressions for area and solve for AD. Ed Gray

Found out BC by Cos Law. Repeatedly used Sin Law to get angle DCA and hence angle ADC; then with AC given got AD. It is just the normal way. Karthik Venkata's formula is really very helpful.