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

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Answer 1 · 2018-02-17T09:06:15

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

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

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