Express Lengths As Function of theta

David Cohen

Just do part (a). I will parts (b & c).

 

a.
BC is hypothenuse

sin(theta)=5/BC

BC =5/sin(theta)

 

Let me see.

Part (b)

For the little triangle above we use cosine.

cos(theta) = 4/AB

AB•cos(theta) = 4

AB = 4/cos(theta)

Part (c)

Here we use the sine function.

From A to the right angle inside the little triangle, I will call that distance x. From that right angle to the one directly beneath, the distance is 5.

sin(theta) = (x + 5)/(AB + BC)

Note AB + BC = AC.

sin(theta) = (x + 5)/AC

AC•sin(theta) = (x + 5)

AC = (x + 5)/sin(theta)

Yes?