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.

We are given with the following value:
AB = BC =90 cm
∠B =25°
∠D=90°

Solving the ∠C:
180 ° = (∠B/2) + ∠D +∠C
180° = (25°/2) + 90° + ∠C
180° - 12.5° -90°  = ∠C
∠C = 77.5°

Solving for length in mD:
sin 77.5° = mD/90
mD = 90 sin77.5°
mD = 87.87

Solving for mDC:
cos 77.5 = mDC/90
mDC = 90 cos 77.5°
mDC = 19.48

Area = 1/2 * Base * Height
Area = 1/2 * 19.48*87.87
Area = 855.834

The total area of triangular pennant = 2 x 855.834 = 1,711.67 cm².

The area is 1,711.67 cm².