How to get the angle in the triangle

Question

See the following pictures: ∠DBC=13°,∠BCD=30°,∠DCA=73°,∠CAB=51°

∠ADB=?

Answer 1

Let $\measuredangle ADB=x$.

Thus, by law of sines we obtain: $$\frac{AD}{AB}=\frac{\sin\measuredangle DBA}{\measuredangle ADB}=\frac{\sin13^{\circ}}{\sin{x}},$$ $$\frac{AC}{AD}=\frac{\sin\measuredangle ADC}{\sin\measuredangle ACD}=\frac{\sin(223^{\circ}-x)}{\sin73^{\circ}}.$$ Thus, $$\frac{AC}{AB}=\frac{\sin13^{\circ}\sin(223^{\circ}-x)}{\sin{x}\sin73^{\circ}},$$ which gives $$\frac{\sin13^{\circ}\sin(223^{\circ}-x)}{\sin{x}\sin73^{\circ}}=\frac{\sin26^{\circ}}{\sin103^{\circ}}$$ or $$\frac{\sin13^{\circ}\sin(223^{\circ}-x)}{\sin{x}\sin73^{\circ}}=\frac{2\sin13^{\circ}\cos13^{\circ}}{\cos13^{\circ}}$$ or $$\frac{\sin(x-43^{\circ})}{\sin{x}\sin73^{\circ}}=2$$ or $$\frac{\sin{x}\cos43^{\circ}-\cos{x}\sin43^{\circ}}{\sin{x}}=2\sin73^{\circ}$$ or $$\cos43^{\circ}-\sin43^{\circ}\cot{x}=2\sin73^{\circ}$$ or $$\cot{x}=\frac{\cos43^{\circ}-2\sin73^{\circ}}{\sin43^{\circ}}.$$ Now, $$\frac{\cos43^{\circ}-2\sin73^{\circ}}{\sin43^{\circ}}=\frac{\cos43^{\circ}-\cos17^{\circ}-\cos17^{\circ}}{\sin43^{\circ}}=$$ $$=\frac{-2\sin13^{\circ}\sin30^{\circ}-\cos17^{\circ}}{\sin43^{\circ}}=-\frac{\sin13^{\circ}+\sin73^{\circ}}{\sin43^{\circ}}=$$ $$=-\frac{2\sin43^{\circ}\cos30^{\circ}}{\sin43^{\circ}}=-\sqrt3,$$

which gives $x=150^{\circ}$.

Done!

Answer 2

Pls see the picture as follows:

Assume $\overline{CD}=2$, we have: $$\frac{sin\alpha}{sin\beta}=2\bullet cos17^{\circ}$$

But: $$\alpha+\beta=51^{\circ}=3\bullet 17^{\circ}$$

So, we have: $\alpha=2\beta=2\times 17^{\circ}$, which is the only solution.

because the equation :$sin\alpha=2cos17^{\circ}\cdot sin(51^{\circ}-\alpha)$, left side is increasing function and right side is decreasing function, so there is only ONE solution in $(0,51^{\circ})$.

which gives $\theta=180^{\circ}-(13^{\circ}+17^{\circ}) = 150^{\circ}$