Atriangular pennant has two sides that are 90 cm long each with an included angle of 25°.
what is the area of this pennant?

enter your answer as a decimal in the box. round only your final answer to the nearest tenth.
? cm^2

See attached picture for reference of the problem.
We have angle A = 25°
We have two sides equal to 90 cm.
Then we can solve for the height "h" such as shown below:
cos A/2 = h /90
cos 25/2 = h/90
cos 12.5 = h/90
h = 90*cos12.5
h = 87.87 cm

Solving for the base:
sin 12.5  = (b/2)/90
b/2 = 90sin12.5
b=38.96 cm

Solving for the area (it is an isosceles triangle since two sides are equal:
Area = b*h/2
Area = 38.96*87.87 /2
Area = 1,711.67 cm²