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Answer :

Answer:

Use the law of cosines to find c:

  • c = [tex]\sqrt{a^2 + b^2 - 2ab*cos C}[/tex]
  • c = [tex]\sqrt{2^2+6^2-2*2*6*cos 95}[/tex] = [tex]\sqrt{42.09}[/tex] = 6.49

Use the low of sines to find one of the missing angles:

  • sin A / 2 = sin 95° / 6.49
  • A = arcsin (2 sin 95° / 6.49)
  • A = 17.9°

Find the missing angle:

  • B = 180° - (A + C)
  • B = 180° - (95° + 17.9°) = 67.1°

Answer:

c ≈ 6.49

A ≈ 17.9°

B ≈ 67.1°

Step-by-step explanation:

To find the length of side c, use the Cosine rule.

Cosine rule

[tex]\sf c^2=a^2+b^2-2ab \cos C[/tex]

(where a, b and c are the sides and C is the angle opposite side c)

From inspection of the given triangle:

  • a = 2
  • b = 6
  • C = 95°

Substitute the given values into the formula and solve for c:

[tex]\implies \sf c^2=2^2+6^2-2(2)(6) \cos 95^{\circ}[/tex]

[tex]\implies \sf c^2=4+36-24 \cos 95^{\circ}[/tex]

[tex]\implies \sf c=\sqrt{40-24 \cos 95^{\circ}}[/tex]

[tex]\implies \sf c=6.49\:\:(2\:d.p.)[/tex]

To find angles A and B, use the Sine Rule.

Sine Rule

[tex]\sf \dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}[/tex]

(where A, B and C are the angles and a, b and c are the sides opposite the angles)

From inspection of the given triangle:

  • a = 2
  • b = 6
  • [tex]\sf c=\sqrt{40-24 \cos 95^{\circ}}[/tex]
  • C = 95°

Substitute the given values into the formula:

[tex]\implies \sf \dfrac{\sin A}{2}=\dfrac{\sin B}{6}=\dfrac{\sin 95^{\circ}}{\sqrt{40-24 \cos 95^{\circ}}}[/tex]

Solving for A:

[tex]\implies \sf \dfrac{\sin A}{2}=\dfrac{\sin 95^{\circ}}{\sqrt{40-24 \cos 95^{\circ}}}[/tex]

[tex]\implies \sf A=\sin^{-1}\left(\dfrac{2\sin 95^{\circ}}{\sqrt{40-24 \cos 95^{\circ}}}\right)[/tex]

[tex]\implies \sf A=17.9^{\circ}\:\:(1\:d.p.)[/tex]

Solving for B:

[tex]\implies \sf \dfrac{\sin B}{6}=\dfrac{\sin 95^{\circ}}{\sqrt{40-24 \cos 95^{\circ}}}[/tex]

[tex]\implies \sf B=\sin^{-1}\left(\dfrac{6\sin 95^{\circ}}{\sqrt{40-24 \cos 95^{\circ}}}\right)[/tex]

[tex]\implies \sf B=67.1^{\circ}\:\:(1\:d.p.)[/tex]

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