Find the area of an isosceles triangle with a side of 16cm and an angle of 15 degrees at the base.

ΔАВС – isosceles.

AB = 16 cm.

∠CAB = 15 “.

S (ΔАВС) -?

The area of ​​a triangle S is determined by the formula: S = a * h / 2, where a is the side of the triangle (base), to which the height h is drawn.
From the top B, let us lower the height BD to the base of the AC.
Point D divides the base into equal segments AD = DC.

The area of ​​the triangle S (ΔАВС) will be determined by the formula: S (ΔАВС) = АС * ВD / 2 = 2 * АD * ВD / 2 = АD * ВD.
Consider a right-angled triangle ΔАВD.
By definition sin∠CAB = BD / AB, cos∠CAB = AD / AB.

BD = AB * sin∠CAB, AD = AB * cos∠CAB.

S (ΔABS) = AB * cos∠CAB * AB * sin∠CAB = AB ^ 2 * cos∠CAB * sin∠CAB.

S (ΔABS) = (16 cm) ^ 2 * cos15 “* sin∠15” = 256 cm ^ 2 * 0.966 * 0.259 = 64 cm ^ 2.

Answer: the area of ​​the triangle is S (ΔABS) = 64 cm ^ 2.