VIDEO ANSWER: Solve the triangle A B C, given a=525, c=421, and A=130^{\circ} 50^{\prime}. Refer to Fig. 31-9. Since A is obtuse and a>c, there is one solution. For C: \quad \sin C=\frac{c \sin A}{a}=\frac{421 \sin
Solve the triangle A B C, given a=525, c=421, and A=130^∘ 50^'. Refer to Fig. 31-9. Since A is obtuse and a>c, there is one solution. For C: sinC=(c sinA)/(a)=(421 sin130^∘ 50^')/(525)=(421(0.7566))/(525)=0.6067 and C=37^∘ 20^' For B : B=180^∘-(C+A)=11^∘ 50^' For b : b=(a sinB)/(sinA)=(525 sin11^∘ 50^')/(sin130^∘ 50^')=(525(0.2051))/(0.7566)=142 The required parts are C=37^∘ 20^', B=11^∘ 50^', and b=142. Fig. 31-9 (FIGURE CAN'T COPY)
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M13 - Solution of Triangles, PDF, Sine
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M13 - Solution of Triangles, PDF, Sine
M13 - Solution of Triangles, PDF, Sine
