The perpendicular to the plane is 16 cm. Find the inclined and its projection onto the plane

The perpendicular to the plane is 16 cm. Find the inclined and its projection onto the plane if the angle between the inclined and perpendicular is 60 degrees.

It is necessary to find the inclined AB and its projection AC on the plane S.
Since AC ┴ BC, ABC is a right-angled triangle.
According to the condition of the assignment, ∠ABS = 60 °. Then from ∠А + ∠В + ∠С = 180 ° we have ∠А = 180 ° – ∠В – ∠С = 180 ° – 60 ° – 90 ° = 30 °.
The leg, lying opposite an angle of 30 degrees, is equal to half of the hypotenuse, that is, BC = AB: 2, whence AB = 32 cm.
By the Pythagorean theorem: AB ^ 2 = AC ^ 2 + BC ^ 2 or AC ^ 2 = AB ^ 2 – BC ^ 2 = (32 cm) ^ 2 – (16 cm) ^ 2 = 768 cm2, whence AC = 16√ 3 cm.
Answer: 32 cm; 16√3 cm.