How do you solve the triangle given #C=15^circ15', a=6.25, b=2.15#?

1 Answer
Jan 29, 2017

Answer:

Covert minutes to decimal degrees
Use The Law of Cosines to find the length of side c
Use The Law of Sines to find #B#
Use the sum of the interior angles equals #180^@# to find #A#

Explanation:

Use the factor #1^@/(60')# to convert the minutes to decimal degrees:

#(15')/1(1^@)/(60')= (15/60)^@ = 0.25^@#

#C = 15.25^@#

Use The Law of Cosines to find the length of side c:

#c=sqrt(a^2+b^2-2(a)(b)cos(C)#

#c=sqrt((6.25)^2+(2.15)^2-2(6.25)(2.15)cos(15.25^@)#

#c~~4.21#

Use The Law of Sines to find B:

#sin(B)/b = sin(C)/c#

#B = sin^-1(sin(C)b/c)#

#B = sin^-1(sin(15.25^@)2.15/4.21)#

#B~~ 7.72^@#

To find A use the some if the interior angle equal #180^@#:

#A + B + C = 180^@#

#A + 7.72^@ + 15.25^@ = 180^@#

#A = 157.03^@#