How do you solve the triangle given #B=2^circ45', b=6.2, c=5.8#?

1 Answer
Feb 17, 2018

Thus, the solution is

#a=12.271, b=6.2, c=5.8# form the sides, while
#A=174^@34', B=2^@45', C=2^@41'# form the angles

Explanation:

By the sine rule

#a/sinA=b/sinB=c/sinC#
Given:

#B=2^@45', b=6.2, c=5.8#

We need to find the angle C immediately

#b/sinB=c/sinC#

#6.2/sin2^@45'=5.8/sinC#
Since the angles are small, #<5^@'#, #sintheta=theta#

when theta is expressed in radians

#2^@45'=2^@+(45/60)^@=2.75^@#

#2.75^@=pi/180xx2.75^@=0.048#

Thus, #sin2^@45'=sin0.05rad=0.048#
and

Thus, #6.2/0.048=5.8/sinC#

#sinC=5.8xx0.048/6.2=0.047#

#C=0.047rad=0.047xx180/pi=2.68^@#

#0.68^@=0.68xx60 min=41'#
hence, #C=2^@41'#

To find a
#sinA=sin(B+C)# by allied angles

#sin(B+C)=sin(2^@45'+2^@41')=sin5.433#
#sinA=sin0.095rad=0.095#

#a/sinA=b/sinB#

#a/0.095=6.2/sin2^@45'=6.2/sin0.048#
#a=602xx0.095/0.048=12.271#
#A+B+C=180#
#A+2^@45'+2^@41'=180#

#A=174^@34'#

Thus, the solution is

#a=12.271, b=6.2, c=5.8# form the sides, while
#A=174^@34', B=2^@45', C=2^@41'# form the angles