Question #236237

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


(Hint: find the other sides and the area)


1
Expert's answer
2021-09-17T03:54:27-0400

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

B=70°\therefore \angle B=70\degree

Now using sine law, we get:

sinAa=sinBb=sinCcsin66°7=sin70°b=sin44°c0.917=0.93b=0.69cb=7.15cm; c=5.3cm\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 ABC=12bcsinA\triangle ABC=\frac{1}{2}bcsinA\\

=12×(7.15)×(5.3)×(sin66°)=12×(7.15)×(5.3)×(0.91)=17.24 cm2=\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


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