Calculate the area of an isosceles triangle ABC with base AC if the angle c is 75 degrees and the side BC is 16 cm.

The angles at the base of an isosceles triangle are:

<C = <A = 75 °.

The sum of the angles of a triangle is 180 °:

<A + <B + <C = 180 °;

75 ° + <B + 75 ° = 180 °;

<B = 180 ° – 75 ° – 75 ° = 30 °.

Let’s lower the height АD to the side AB. The resulting triangle is rectangular. Side AD lies opposite an angle of 30 ° and is equal to half of the hypotenuse BC:

AD = BC / 2 = 16/2 = 8 cm.

In an isosceles triangle with base AC, side AB = BC = 16 cm.

The area of triangle ABC is half the product of side AB and height BC:

S = (AB * BC) / 2 = (16 cm * 8 cm) / 2 = 64 cm ^ 2.

Answer: 64 cm ^ 2.