$$A=a·b·\sin \alpha$$

Based on the parallelogram theorem:

$$2a^2+2b^2={d_1}^2+{d_2}^2$$

$$2b^2={d_1}^2+{d_2}^2-2a^2$$

$$b^2=\frac{{d_1}^2+{d_2}^2-2a^2}{2}$$

$$b^2=\frac{{26}^2+{30}^2-2·{14}^2}{2}$$

$$b^2=\frac{676+900-2·196}{2}=\frac{1184}{2}=592$$

$$b^2=16·37$$

$$b=\sqrt{16·37}=\sqrt{16}·\sqrt{37}$$

The law of cosine can be written for the triangle ABC.

$${d_2}^2=a^2+b^2-2·a·b·\cos \alpha$$

$$\cos \alpha =\frac{a^2+b^2-{d_2}^2}{2ab}$$

$$\cos \alpha =\frac{{14}^2+{\left(4\sqrt{37}\right)}^2-{\left(30\right)}^2}{2·14·4\sqrt{37}}$$

$$\cos \alpha =\frac{196+592-900}{112\sqrt{37}}$$

$$\cos \alpha =\frac{-112}{112\sqrt{37}}$$

$$\cos \alpha =\frac{-1}{\sqrt{37}}$$

$$\sin \alpha =\sqrt{1-{\cos }^2\alpha }$$

$$\sin \alpha =\sqrt{1-{\left(\frac{1}{\sqrt{37}}\right)}^2}$$

$$\sin \alpha =\sqrt{1-\frac{1}{37}}=\sqrt{\frac{37-1}{37}}=\sqrt{\frac{36}{37}}$$

$$A=a·b·\sin \alpha =14·4·\cancel{\sqrt{37}}·\frac{6}{\cancel{\sqrt{37}}}$$