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Given ∆ABC with Angle A = 52°, side c = 29, and a = 27. There are two triangles possible.Find Angle C and C'. Round your answer to the nearest degree.

Respuesta :

To find < C, we will use the sine rule

[tex]\frac{\sin A}{a}=\frac{\sin C}{c}[/tex][tex]\frac{\sin52}{27}=\frac{\sin C}{29}[/tex]

cross-mulply

27sinC = 29sin52

[tex]\sin C=\frac{29\sin 52}{27}[/tex][tex]\sin C=0.84638[/tex]

Take the arcsin of both-side

[tex]C\approx58^o[/tex]

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