How do i solve this calculus problem using what is given?

Question

In the diagram at right, CD is the altitude from C. (a) Express CD in terms of angle B and side a. (b) Express BD in terms of angle B and side a. (c) Simplify the expression (a sin B)2 + (a cos B)2 and discuss its relevance to the diagram. B D А (d) Why was a sin B used instead of sin Ba?

the diagram shows a triangle with the points b c and A and d is made from a straight 90 degree line coming from point c the top and ending between b and A. B at the left corner of the triangle and A at the right. And a is the edge between b and c.

Answer 1

So if I get this right you have a triangle ABC with altitude CD where D lie on the base of the triangle AB and perpendicular to it, you are given side a which is edge CB and want CD

1- Well in this case triangle BCD is a right angle triangle with angle D at 90° and with angle B at left corner and side BC = a. Normally side a is opposite to angle D which which is also the hypotenuse of triangle BCD . From trigonometry sin(B) = CD/ BC, and CD = BC sin(B)= a sin B, then CD = a sin(B)

2- BD is the base of the right angle tringle BCD and cos(B) = BD/BC , BD = BC cos(B) = a cos(B), and

BD = a cos(B)

3- [a sin(B)]²+[a cos(B)]² , since a sin(B) = CD and a cos(B) = BD, then CD²+ BD² = BC² = a²

To prove this [a sin(B)]²+[a cos(B)]² = a²sin(B)² +a²cos(B)² = a² [ sin(B)² + cos(B)²] = a²(1) = a²

4- B is the angle at B and a is side BC . You cannot take sin(aB) because aB is a multiplication of a side in unit length by an angle and that is impossible to take the sin of