Answer to Question #236237 in Trigonometry for luka

Question

Solve the triangle ABC given that A=66 degrees,C=44 degrees and a = 7cm.

(Hint: find the other sides and the area)

Answer 1

Given that $\angle A=66\degree,\angle C=44\degree,a=7 cm$

$\therefore \angle B=70\degree$

Now using sine law, we get:

$\frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}\\ \Rightarrow \frac{sin66\degree}{7}=\frac{sin70\degree}{b}=\frac{sin44\degree}{c}\\ \Rightarrow \frac{0.91}{7}=\frac{0.93}{b}=\frac{0.69}{c}\\ \therefore b=7.15cm;\ c=5.3 cm$

Now, area of $\triangle ABC=\frac{1}{2}bcsinA\\$

$=\frac{1}{2}\times (7.15)\times (5.3)\times (sin66\degree)\\ =\frac{1}{2}\times (7.15)\times (5.3)\times (0.91)\\ =17.24 \ cm^2$