Given:
In triangle ABC, AB=8.2 cm, C=13.5 cm and angle A= 81 degrees.
To find:
The length of AC.
Solution:
According to the Law of Sines:
[tex]\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}[/tex]
Using Law of Sines, we get
[tex]\dfrac{\sin A}{a}=\dfrac{\sin C}{c}[/tex]
[tex]\dfrac{\sin (81^\circ)}{13.50}=\dfrac{\sin C}{8.2}[/tex]
[tex]\dfrac{\sin (81^\circ)\times 8.2}{13.50}=\sin C[/tex]
[tex]\sin^{-1}\dfrac{\sin (81^\circ)\times 8.2}{13.50}= C[/tex]
[tex]C\approx 36.86^\circ[/tex]
Using angle sum property, we get
[tex]\angle A+\angle B+\angle C=180^\circ[/tex]
[tex]81^\circ+\angle B+36.86^\circ=180^\circ[/tex]
[tex] \angle B =180^\circ-81^\circ-36.86^\circ [/tex]
[tex] \angle B =62.14^\circ [/tex]
Now,
[tex]\dfrac{\sin A}{a}=\dfrac{\sin B}{b}[/tex]
[tex]\dfrac{\sin (81^\circ)}{13.50}=\dfrac{\sin (62.14^\circ)}{b}[/tex]
[tex]b=\dfrac{13.50\times \sin (62.14^\circ)}{\sin (81^\circ)}[/tex]
[tex]b\approx 12.08[/tex]
Therefore, the length of AC is 12.08 cm.
