Help trig q

Alevelhelp.1

Help, I don’t get this question can someone walk me through it plz

(edited 1 year ago)

Reply 1

mqb2766

dcos(theta) is the x coordinate of OP, then the other part is the x coordinate of PQ, so add them together to get the x coord of OQ. As the ms suggests, a sketch is useful with the angle(s) for PQ clearly marked on.

(edited 1 year ago)

Reply 2

lordaxil

Think about resolving the lengths of the two rods OP and PQ along the horizontal direction. The horizontal length of OP is just

d \cos(\theta)

. Then, since the horizontal length of PQ is always going backwards relative to that of PQ (for domain of angle

0 \le \theta \le \pi/2

) then its length is

-d \sin (\pi/2 - 2\theta) = d \sin (2\theta - \pi/2)

. To get the angle for the sin term, I drew a right-angled triangle OPA, where A is the intersection of perpendicular from P to x-axis. The transformation is valid since sine is an odd function.