Question

If psina = qsin(120+a) = rsin(240+a), prove that pq+qr+rp = 0​

Answer 1

Answer:

sorry it is tough as I'm in junior class

Answer 2

psina=qsin(120∘+a)=rsin(240∘+a)

⟹psina=qsin[180∘−(60∘−a)]=rsin[180∘+60∘+a)

⟹psina=qsin(60∘−a)=−rsin(60∘+a)

⟹psina=3–√2qcosa−12qsina=−3–√2rcosa−12rsina

psina=3–√2qcosa−12qsina

⟹rpsina=3–√2qrcosa−12qrsina(1)

psina=−3–√2rcosa−12rsina

⟹pqsina=−3–√2qrcosa−12qrsina(2)

by (1) + (2)

(rp+pq)sina=−qrsina

⟹rp+pq=−qr

⟹rp+pq+qr=0

Hope it helps.....