Answer: D
Step-by-step explanation:
By the cosine rule, [tex]BC=\sqrt{10^{2}+14^{2}-2 \cdot 10 \cdot 14 \cdot \cos 70^{\circ}}[/tex]
Now, we can use the law of sines to get that:
[tex]\frac{\sqrt{10^{2}+14^{2}-2\cdot 10 \cdot 14 \cdot \cos 70^{\circ}}}{\sin 70^{\circ}}=\frac{14}{\sin \angle ABC}[/tex]
Meaning that [tex]\sin \angle ABC =\frac{14\sin 70^{\circ}}{\sqrt{10^{2}+14^{2}-2 \cdot 10 \cdot 14 \cdot \cos 70^{\circ}}}[/tex]
And thus [tex]\angle ABC=\sin^{-1} \left(\frac{14\sin 70^{\circ}}{\sqrt{10^{2}+14^{2}-2 \cdot 10 \cdot 14 \cdot \cos 70^{\circ}}} \right) \approx 68.39^{\circ}[/tex]
