Solve the triangle ABC given that A=66 degrees,C=44 degrees and a = 7cm.
(Hint: find the other sides and the area)
Given that ∠A=66°,∠C=44°,a=7cm\angle A=66\degree,\angle C=44\degree,a=7 cm∠A=66°,∠C=44°,a=7cm
∴∠B=70°\therefore \angle B=70\degree∴∠B=70°
Now using sine law, we get:
sinAa=sinBb=sinCc⇒sin66°7=sin70°b=sin44°c⇒0.917=0.93b=0.69c∴b=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 cmasinA=bsinB=csinC⇒7sin66°=bsin70°=csin44°⇒70.91=b0.93=c0.69∴b=7.15cm; c=5.3cm
Now, area of △ABC=12bcsinA\triangle ABC=\frac{1}{2}bcsinA\\△ABC=21bcsinA
=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=21×(7.15)×(5.3)×(sin66°)=21×(7.15)×(5.3)×(0.91)=17.24 cm2
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