In an isosceles triangle ABC – AC – base = 12 cm Angle ABC = 120 degrees Find: a) the height drawn

In an isosceles triangle ABC – AC – base = 12 cm Angle ABC = 120 degrees Find: a) the height drawn to the base AC; b) side

In order to find the length of the ВН height, consider the AВН triangle. In an isosceles triangle, the height that goes out and the angle at the apex divides it in half, and also divides the base of this triangle in half. Thus, the degree measure of the angle ∠В will be equal to:

∠В = 120/2 = 60 °;

AH = AC / 2 = 12/2 = 6 cm.

Now, using the tangent of the angle, we can find the length of the HВ height:

tg B = AH / BH;

BH = AH / tg B;

tg 60 ° = 1.7321;

BH = 6 / 1.7321 = 3.46 cm.

Now, behind the Pythagorean theorem, we find the side AB:

AB ^ 2 = BH ^ 2 + AH ^ 2;

AB ^ 2 = 3.46 ^ 2 + 6 ^ 2 = 11.97 + 36 = 47.97;

AB = √47.97 = 6.93 cm

Answer: The height of the triangle is 3.46 cm, the side is 6.93 cm.