In an isosceles triangle, the outer angle at the apex opposite to the base is 130 degrees. Find the angles of this triangle 2.

In an isosceles triangle, the outer angle at the apex opposite to the base is 130 degrees. Find the angles of this triangle 2. In the right-angled triangle ACB (angle C = 90 degrees) the height CD is drawn. Hypotenuse AB = 10cm, angle CBA = 30g. Find BD.

1). Angle B – the angle at the apex opposite to the base is 180 ° – 130 ° = 50 °;

The angles at the base of an isosceles triangle A and C are equal, which means

<A = <C = (180 ° – 50 °): 2 = 65 °;

2). In a right-angled triangle AB – hypotenuse, AB = 10 cm; <B = 30 °;

AC – leg, lying opposite an angle of 30 °, then AC = 1/2 AB;

AC = 1/2 * 10 = 5 (cm);

By the Pythagorean theorem: BC² = AB² – AC²; BC² = 100 – 25 = 75 (²); BC = √75 (cm);

In a right-angled triangle BCD CD is a leg opposite to an angle of 30 °, CB is a hypotenuse;

means СD = 1/2 CB = 1/2 * √75;

BD² = BC² – CD²; BD² = 75 – 1/4 * 75; BD² = 3/4 * 75; BD = √ (3 * 3 * 25/4);

BD = (3 * 5) / 2; BD = 7.5 cm.