Solve triangle with angles?

with A=40 deg. C=25 deg. b=10

1 Answer
Apr 11, 2018

Solution: Sides are #a~~7.09 , b=10 , c~~ 4.66# unit and
angles are
# /_A=40^0 , /_B=115^0 , /_C= 25^0#

Explanation:

Angle between Sides # b and c# is # /_A=40^0#

Angle between Sides # a and b# is # /_C=25^0 #

Angle between Sides # c and a# is # /_B= 180-(40+25)=115^0#

The sine rule states if #a, b and c# are the lengths of the sides

and opposite angles are #A, B and C# in a triangle, then:

#a/sinA = b/sinB=c/sinC ; b=10 :. b/sinB=c/sinC#

#10/sin115=c/sin25 or c= 10* (sin25/sin115) ~~ 4.66 (2dp) #

Similarly #a/sinA=b/sinB # or

#a/sin 40=10/sin115 or a= 10* (sin40/sin115) ~~ 7.09 (2dp) #

Solution: Sides are #a~~7.09 , b=10 , c~~ 4.66# unit and

angles are # /_A=40^0 , /_B=115^0 , /_C= 25^0# [Ans]