An alpha plane is drawn through the leg of an isosceles right-angled triangle. The angle between the planes

An alpha plane is drawn through the leg of an isosceles right-angled triangle. The angle between the planes of the triangle and the alpha is 60 ‘. Calculate the lengths of the projections of the sides of this triangle on the alpha plane, if the length of the leg of this triangle is 10 dm.

According to the condition, the triangle ABC is isosceles, AC = BC = 10 dm. The angle ВСH is equal to 60. Consider a right-angled triangle ВHС, in which the angle Н is a straight line, since point H is the projection of point B onto the plane α.

Then the angle CBH = 180 – 90 – 60 = 30, the leg CH lies opposite the angle 30, and will be equal to half the length of the hypotenuse BC. CH = BC / 2 = 10/2 = 5 dm.

Then BH / CB = Sin60. ВH = CB * Sin60 = (10 * √3) / 2 = 5 * √3 dm.

Consider a right-angled triangle ABC, in which AC = BC = 10 dm, then, according to the Pythagorean theorem, AB ^ 2 = AC ^ 2 + BC ^ 2 = 100 + 100 = 200.

AB = √200 = 10 * √2 dm.

Consider a right-angled triangle ACH, in which the angle C is straight, and by the Pythagorean theorem we define the leg AH.

AH ^ 2 = AC ^ 2 – CH ^ 2 = (10 * √2) ^ 2 – 5 ^ 2) = 200 – 25 = 125.

AH = √125 = 5 * √5 dm.

Answer: CH = 5 dm, AH = 5 * √5 dm.