Find the perimeter of each of the two noncongruent triangles where a = 15, b = 20, and A = 29°
Accepted Solution
A:
Answer with explanation:Using Sine Rule for Congruence of Triangles [tex]\Rightarrow\frac{a}{\ SinA}=\frac{b}{\ Sin B}=\frac{c}{\ Sin C}\\\\\Rightarrow\frac{15}{\ Sin29^{\circ}}=\frac{20}{\ Sin B}\\\\\Rightarrow\frac{15}{0.49}=\frac{20}{\ Sin B}\\\\\Rightarrow \ SinB=\frac{20 \times 0.49}{15}\\\\\Rightarrow \ SinB=\frac{9.8}{15}\\\\\Rightarrow \ SinB=0.65\\\\B=41^{\circ}[/tex]Using Angle Sum Property of Triangle⇒∠A+∠B+∠C=180°⇒29°+41°+∠C=180°⇒∠C=180°-70°⇒∠C=110°→Again Using Sine Rule[tex]\Rightarrow \frac{b}{\ Sin B}=\frac{c}{\ Sin C}\\\\\Rightarrow \frac{20}{\ Sin 41^{\circ}}=\frac{c}{\ Sin 110^{\circ}}\\\\\Rightarrow \frac{20}{0.65}=\frac{c}{0.94}\\\\\Rightarrow \frac{20 \times 0.94}{0.65}=c\\\\\Rightarrow c=\frac{18.8}{0.65}\\\\\Rightarrow c=28.92[/tex]Length of third Side =28.92 unitSo,Perimeter of Triangle =Sum of sides of triangle=a +b +c =15 + 20 +28.92 = 63.92 unit