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  • 20-12-2022
  • Mathematics
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Solve the triangle PLEASEEEEEE

Solve the triangle PLEASEEEEEE class=

Respuesta :

Medunno13
Medunno13 Medunno13
  • 20-12-2022

Part 1

[tex]a=\sqrt{12^2 + 21^2 -2(12)(21) \cos 35^{\circ}} \approx \boxed{13}[/tex]

Part 2

[tex]\frac{\sin B}{b}=\frac{\sin A}{a}\\\\\sin B=\frac{b\sin A}{a}\\\\\sin B=\frac{12 \sin 35^{\circ}}{\sqrt{12^2 +21^2 -2(12)(21) \cos 35^{\circ}}}\\\\B=\sin^{-1} \left(\frac{12\sin 35^{\circ}}{\sqrt{12^2 +21^2 -2(12)(21) \cos 35^{\circ}}} \right)\\\\B \approx \boxed{32^{\circ}}[/tex]

Part 3

[tex]\angle C=180^{\circ}-\angle A -\angle B=\boxed{113^{\circ}}[/tex]

Answer Link

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