A triangle has two corners with angles of # pi / 4 # and # pi / 6 #. If one side of the triangle has a length of #3 #, what is the largest possible area of the triangle?

Question

A triangle has two corners with angles of #  pi  /  4  # and #  pi  /  6   #.  If one side of the triangle has a length of #3  #, what is the largest possible area of the triangle?

Sophia Alfred · Accepted Answer

Largest possible area of triangle is #6.15# sq.unit

Angle between Sides # A and B# is # /_c= pi/4=45^0# Angle between Sides # B and C# is # /_a= pi/6=30^0 :.# Angle between Sides # C and A# is # /_b= 180-(45+30)=105^0# Length of one side is #3# , For largest area of triangle #3# should be smallest side , which is opposite to the smallest angle #/_a=30^0 :. A=3# 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/sin a = B/sin b=C/sin c ; A=3# # :. A/sin a=B/sin b or 3/sin 30 = B/sin 105# or #B= 3* sin 105/sin 30 ~~ 5.8# Now we know sides #A=3 , B=5.8# and their included angle #/_c =45^0#. Area of the triangle is #A_t=(A*B*sin c )/2# #:. A_t=(3*5.8*sin 45 )/2 ~~ 6.15# sq.unit Largest possible area of triangle is #6.15# sq.unit [Ans]

Wyatt Anderson · Answer

undefined The largest possible area of the triangle can be calculated using the formula for the area of a triangle, which is $ \frac{1}{2} \times \text{base} \times \text{height} $. Using the given angles, we can determine the lengths of the other two sides of the triangle using trigonometric ratios. The side opposite the $ \frac{\pi}{6} $ angle is $ 3\sin(\frac{\pi}{6}) = \frac{3\sqrt{3}}{2} $ units. The side opposite the $ \frac{\pi}{4} $ angle can be found similarly. Then, we can use the formula for the area of a triangle to find the largest possible area.