Answer:
Sum of interior angles of a triangle = 180°
Sine rule to find side lengths:
[tex]\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}[/tex]
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Question 1a
Given: A = 40°, B = 20°, a = 2
40° + 20° + C = 180°
⇒ C = 120°
[tex]\dfrac{2}{\sin 40}=\dfrac{b}{\sin 20}=\dfrac{c}{\sin 120}[/tex]
[tex]\implies b=\sin20 \cdot\dfrac{2}{\sin 40}=1.064177772...[/tex]
[tex]\implies c=\sin 120 \cdot \dfrac{2}{\sin 40}=2.694592711...[/tex]
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Question 2b
Given: A = 50°, C = 20°, a = 3
50° + 20° + B = 180°
⇒ B = 110°
[tex]\dfrac{3}{\sin 50}=\dfrac{b}{\sin 110}=\dfrac{c}{\sin 20}[/tex]
[tex]\implies b=\sin 110 \cdot\dfrac{3}{\sin 50}=3.680044791...[/tex]
[tex]\implies c=\sin 20 \cdot\dfrac{3}{\sin 50}=1.339426765...[/tex]
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Question 3c
Given: B = 70°, C = 10°, b = 5
70° + 10° + A = 180°
⇒ A = 100°
[tex]\dfrac{a}{\sin 100}=\dfrac{5}{\sin 70}=\dfrac{c}{\sin 10}[/tex]
[tex]\implies a=\sin 100 \cdot \dfrac{5}{\sin 70}=5.240052605...[/tex]
[tex]\implies c=\sin 10 \cdot \dfrac{5}{\sin 70}=0.9239626545...[/tex]
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Question 4d
Given: A = 70°, B = 60°, c = 4
70° + 60° + C = 180°
⇒ C = 50°
[tex]\dfrac{a}{\sin 70}=\dfrac{b}{\sin 60}=\dfrac{4}{\sin 50}[/tex]
[tex]\implies a=\sin 70 \cdot \dfrac{4}{\sin 50}=4.906726388...[/tex]
[tex]\implies b=\sin 60 \cdot \dfrac{4}{\sin 50}=4.522063499...[/tex]
