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 1aa
Given: B = 10°, C = 100°, b = 2
10° + 100° + A = 180°
⇒ A = 70°
[tex]\dfrac{a}{\sin 70}=\dfrac{2}{\sin 10}=\dfrac{c}{\sin 100}[/tex]
[tex]\implies a=\sin 70 \cdot \dfrac{2}{\sin 10}=10.82294826...[/tex]
[tex]\implies c=\sin 100\cdot \dfrac{2}{\sin 10}=11.34256364...[/tex]
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Question 2bb
Given: A = 40°, B = 40°, c = 2
40° + 40° + C = 180°
⇒ C = 100°
[tex]\dfrac{a}{\sin 40}=\dfrac{b}{\sin 40}=\dfrac{2}{\sin 100}[/tex]
[tex]\implies a=\sin 40\cdot\dfrac{2}{\sin 100}=1.305407289...[/tex]
[tex]\implies b=\sin 40\cdot\dfrac{2}{\sin 100}=1.305407289...[/tex]
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Question 1aa
Given: A = 110°, C = 30°, c = 2
110° + 30° + B = 180°
⇒ B = 40°
[tex]\dfrac{a}{\sin 110}=\dfrac{b}{\sin 40}=\dfrac{3}{\sin 30}[/tex]
[tex]\implies a=\sin 110 \cdot \dfrac{3}{\sin 30}=5.638155725...[/tex]
[tex]\implies b=\sin 40\cdot\dfrac{3}{\sin 30}=3.856725658...[/tex]
