How do you solve a triangle given a=6.3. b=9.4, <C=38.3° ?

1 Answer
Nov 18, 2016

Answer:

#"side " c ~~5.9#
#/_A ~~41.23^circ#
#/_B~~100.47^circ#

Explanation:

Here is what I think the triangle looks like:
enter image source here

By the Law of Cosines:
#c=sqrt(a^2+b^2-2ab*cos(C))#
using the given values for #a# and #b#
#c~~5.9246#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

By the Law of Sines:
#color(red)(" ----------------------------------")#
#color(red)("| See note of caution below |")#
#color(red)(" ----------------------------------")#

#sin(A)/a=sin(C)/c#
#rarr sin(A)=(axxsin(C))/c#
# rarr A="arcsin"((axxsin(C))/c)color(white)("XXX")color(red)("see below")#
with given and determined values
#/_A="arcsin"((6.3xxsin(38.3^circ))/5.9246)#
#color(white)("XX") ~~41.23^circ#

#/_B = 180^circ - (/_A+/_C)#
#color(white)("XX")~~100.47^circ#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(red)("CAUTION")#

The conversion of #sin(A)=(axxsin(C))/c#
into #A="arcsin"((axxsin(C))/c)#
is only valid provided angle A is acute

If a triangle has an obtuse angle, that angle will be opposite the longest side (note there can be at most one obtuse angle).
In other words, if an angle is opposite a side which is not the longest side, that angle must be acute.

It is therefore safe to use the Law of Sines to evaluate #/_A# (since side #a < # side #b#).

The largest angle (the angle opposite the longest side) should always be evaluated as #180^circ# minus the sum of the other interior angles, since the Law of Sines may or may not work. (In this particular example, the Law of Sines would have given an incorrect value for #/_B#).