A triangle has sides A, B, and C. Sides A and B have lengths of 5 and 9, respectively. The angle between A and C is #(17pi)/24# and the angle between B and C is # (pi)/8#. What is the area of the triangle?

1 Answer
Jan 19, 2016

area of the triangle# ~~ 12.889" square unites"# to 3 decimal places

Explanation:

Tony B

#color(blue)("Method")#
Find the height h then use: #"area=1/2xxAxxh#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Consider the option to find C and hence h")#

Cosine Rule:
#C^2=A^2+B^2-2ABcos(/_bca)#

Sine Rul:
#C/(sin(/_bca)) = B/(sin(/_abc))=A/(sin(/_bac))#

Both require that the angle #/_bca# is determined.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Using: Sum of internal angles of a triangle# =180^o-> (pi " radians")#

#color(blue)(=> /_bca=pi-(17pi)/24-pi/8 =pi/6 -> (30^o))#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determining length of C")#

Using the Sine Rule (simpler: no square roots!)

#C/(sin(pi/6))=A/(sin(pi/8))#

Known: #color(white)(...)sin(pi/6)=sin(30^o)= 1/2#

#color(brown)(=>C=1/2xx5/(sin(pi/8)) ~~6.499)# to 3 decimal places
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determining height h")#

Project the line cb to d such that a vertical line from 'a' forms 'ad' and #/_bda=pi/2 ->(90^o)#

Then #/_dba= pi-(17pi)/24=(7pi)/24 ->52 1/2 " degrees"#

Using basic trig
#sin((7pi)/24) = h/C#

#=>h=Csin((7pi)/24)#

#color(brown)(h~~ 5.156)# to 3 decimal places
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("To determine area of triangle")#

using: area #= 1/2 xx Axx h#

#color(brown)("area "= 1/2xx 5xx5.156... ~~ 12.889" square unites"# to 3 decimal places