Notice that we have two triangles with the SAME angle, abd with also a common side (the same length) AD.
we can use the law os sines in the smaller triangle and specially using the sides that are known.
for example, we can state in the first (smaller) triangle that:
[tex]\frac{\sin(\theta)}{2}=\frac{\sin (D)}{5.9}=\text{ }\frac{\sin(C)}{AD}[/tex]For the full triangle ABC we have the following law of sines:
[tex]\frac{\sin(2\theta)}{2+\text{?}}=\text{ }\frac{\sin(C)}{8.1}=\frac{\sin (B)}{5.9}[/tex]For the medium triangle ADB the law of sines goes as:
[tex]\frac{\sin(\theta)}{?}=\frac{\sin(B)}{AD}=\frac{\sin(180-D^{})}{8.1}=\frac{\sin (D)}{8.1}[/tex]Now, we need to find common variables to combine equations based on the law of sines.
Notice as well that sin(180-a) = sin(a) this is a trig identity, so we are going to replace this in the last trig identity for triangle ADB
Now, we have the following relationships from the veri first law of sines:
[tex]\sin (\theta)=\frac{2\cdot\sin (D)}{5.9}[/tex]and from the last law of sines we have the folloowing relationship:
[tex]\sin (\theta)=\frac{?\cdot\sin (D)}{8.1}[/tex]so we can equal both sine expressions since they are from the same angle, and try to solve for the unknown "?" in the equation:
[tex]\begin{gathered} \frac{2\cdot\sin(D)}{5.9}=\frac{?\cdot\sin (D)}{8.1} \\ \frac{2}{5.9}=\frac{?}{8.1} \\ \frac{2\cdot8.1}{5.9}=\text{?} \end{gathered}[/tex]whre we have eliminated sin(D) as common factor in both equations (this is correct as long as sin*D) is not equal to zero, which cleary is not the case here)
Therefore our unknown "?" is 2*8.1 / 5.9 = 2.7457 which rounded to one decimal is 2.7
