The height drawn to the base of the isosceles triangle is 8.2 cm, and the side of the triangle is 16.4 cm. Find the corners of this triangle.
Triangle ABC – isosceles, AC – base, BH = 8.2 cm – height, AB = BC = 16.4 cm.
Consider the triangle ABН. AВН is a right-angled triangle, since the height of the ВН is perpendicular to the base of the AC, that is, when it intersects with the AC, it forms an angle of 90 degrees. In the triangle ABН AB is the hypotenuse, since it lies opposite the right angle.
Find the sine of angle A at the base (the sine of the angle is the ratio of the opposite leg to the hypotenuse):
sinA = ВН / AB = 8.2 / 16.4 = 1/2
angle A = 30 degrees.
Since ABC is an isosceles triangle, the angles at its base AC are equal to each other. Then angle A = angle C = 30 degrees.
From the theorem on the sum of the angles of a triangle, it is known that the sum of the angles of any triangle is 180 degrees, that is:
angle A + angle B + angle C = 180 degrees;
30 degrees + angle B + 30 degrees = 180 degrees;
angle B = 180 degrees – 60 degrees;
angle B = 120 degrees.
Answer: angle A = 30 degrees, angle B = 120 degrees, angle C = 30 degrees.
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