To solve for the remaining angles and side of the triangle, we can use the Law of Sines and the fact that the sum of the angles in a triangle is 180 degrees. First, we can find angle C C C using the Law of Sines, then find angle A A A by subtracting the sum of angles B B B and C C C from 180 degrees. Finally, we can use the Law of Sines again to find side a a a.
We are given the following values for the triangle:
Using the Law of Sines, we can find angle C C C: csinC=bsinB \frac{c}{\sin C} = \frac{b}{\sin B} sinCc=sinBb Rearranging gives: sinC=c⋅sinBb \sin C = \frac{c \cdot \sin B}{b} sinC=bc⋅sinB Substituting the known values: sinC=3⋅sin(125∘)6.5 \sin C = \frac{3 \cdot \sin(125^\circ)}{6.5} sinC=6.53⋅sin(125∘) Calculating this yields: C≈22.21∘ C \approx 22.21^\circ C≈22.21∘
Using the fact that the sum of angles in a triangle is 180∘ 180^\circ 180∘: A=180∘−B−C A = 180^\circ - B - C A=180∘−B−C Substituting the values: A=180∘−125∘−22.21∘≈32.79∘ A = 180^\circ - 125^\circ - 22.21^\circ \approx 32.79^\circ A=180∘−125∘−22.21∘≈32.79∘
Using the Law of Sines again to find side a a a: asinA=bsinB \frac{a}{\sin A} = \frac{b}{\sin B} sinAa=sinBb Rearranging gives: a=b⋅sinAsinB a = \frac{b \cdot \sin A}{\sin B} a=sinBb⋅sinA Substituting the known values: a=6.5⋅sin(32.79∘)sin(125∘)≈4.3 a = \frac{6.5 \cdot \sin(32.79^\circ)}{\sin(125^\circ)} \approx 4.3 a=sin(125∘)6.5⋅sin(32.79∘)≈4.3
The remaining angles and side of the triangle are: C≈22.21∘,A≈32.79∘,a≈4.3 C \approx 22.21^\circ, \quad A \approx 32.79^\circ, \quad a \approx 4.3 C≈22.21∘,A≈32.79∘,a≈4.3 Thus, the final answers are: C≈22.21∘,A≈32.79∘,a≈4.3 \boxed{C \approx 22.21^\circ}, \quad \boxed{A \approx 32.79^\circ}, \quad \boxed{a \approx 4.3} C≈22.21∘,A≈32.79∘,a≈4.3
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