How do you solve the triangle given A=40, b=7, c=6?

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Answer 1 · 2017-01-13T19:09:03

#angle B= 81°56'14'', a = 4.544, angle C=58°03'46''#

Cosine rule:

#a^2=b^2+c^2-2bc CosA#

#a^2=(7^2)+(6^2)-2 xx 7 xx 6 xx Cos40°#

#a^2=49+36-84 xx 0.766044443#

#a^2=85-64.34773322#

#a^2=20.65226678#

#a=sqrt20.65226678#

#a=4.544=BC#

#b^2=a^2+c^2-2ac CosB#

# b^2=(4.544)^2+6^2-2 xx 4.544 xx CosB #

# 7^2=(4.544)^2+36-54.528 xx CosB #

#7^2=20.648+36-54.528 xx CosB #

#7^2=56.648-54.528 xx CosB #

#-54.528cosB=49-56.648#

#54.528CosB=56.648-49#

#CosB=7.648/54.528#

# CosB=0.140258216#

#angleB=81° 56’ 14’’#

#180°-(40°+81° 56' 14'')=58° 03' 46'' =angle C#

Check:

#(BC)/(Sin40)=6/(Sin58° 03’ 46’’)#

#BC=(6 xx Sin40°)/(Sin58° 03’ 46’’)#

#BC=3.856725658/0.848628208#

#BC=4.545=a#