Answer:
1. For quadrilateral ABCD:
AB = AD = 2.7, BC = DC = 3.2, and AC = 3.168
<BAC = <DAC [tex]65.5^{o}[/tex], <ABC = <ADC = [tex]64.3^{o}[/tex], and <ACB = <ACD = [tex]50.2^{o}[/tex]
2. For quadrilateral ABCE:
AB = EC = 2.7, BC = AE = 3.2, and AC = 3.168
<BAC = <ACE [tex]65.5^{o}[/tex], <ABC = <AEC = [tex]64.3^{o}[/tex], and <ACB = <EAC = [tex]50.2^{o}[/tex]
Step-by-step explanation:
Applying the Cosine rule to triangle ABC,
[tex]/AC/^{2}[/tex] = [tex]/AB/^{2}[/tex] + [tex]/BC/^{2}[/tex] - 2/AB/ x /BC/ Cos B
= [tex]2.7^{2}[/tex] + [tex]3.2^{2}[/tex] - 2 x 2.7 x 3.2 Cos 64.3
= 7.29 + 10.24 - 17.28 x 0.4337
= 17.53 - 7.49434
= 10.03566
AC = [tex]\sqrt{10.03566}[/tex]
= 3.168
Applying the Sine rule,
[tex]\frac{a}{Sin A}[/tex] = [tex]\frac{b}{Sin B}[/tex] = [tex]\frac{c}{Sin C}[/tex]
So that:
[tex]\frac{3.2}{Sin A}[/tex] = [tex]\frac{3.168}{Sin 643.^{o} }[/tex]
[tex]\frac{3.2}{Sin A}[/tex] = [tex]\frac{3.168}{0.9011}[/tex]
Sin A = [tex]\frac{3.2 * 0.9011}{3.168}[/tex]
= [tex]\frac{2.88352}{3.168}[/tex]
= 0.9102
⇒ A = [tex]Sin^{-1}[/tex] 0.9102
= [tex]65.5^{o}[/tex]
But sum of angles in a triangle is [tex]180^{o}[/tex], so that;
A + B + C = [tex]180^{o}[/tex]
65.5 + 64.3 + C = [tex]180^{o}[/tex]
129.8 + C = [tex]180^{o}[/tex]
C = [tex]180^{o}[/tex] - 129.8
C = [tex]50.2^{o}[/tex]
1. For quadrilateral ABCD:
AB = AD = 2.7, BC = DC = 3.2, and AC = 3.168
<BAC = <DAC [tex]65.5^{o}[/tex], <ABC = <ADC = [tex]64.3^{o}[/tex], and <ACB = <ACD = [tex]50.2^{o}[/tex]
2. For quadrilateral ABCE:
AB = EC = 2.7, BC = AE = 3.2, and AC = 3.168
<BAC = <ACE [tex]65.5^{o}[/tex], <ABC = <AEC = [tex]64.3^{o}[/tex], and <ACB = <EAC = [tex]50.2^{o}[/tex]
