anonymous
  • anonymous
prove that the length of the perpendicular from the origin to the straight line joining the two points having coordinates ( a cos alpha, a sin alpha) and (a cos beta, a sin beta) is a cos { ( alpha + beta)/2}.
Mathematics
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
Equation of line passing through ( a cos alpha, a sin alpha) and (a cos beta, a sin beta) is y-a sin alpha = (a sinbeta -a sinalpha)/( a cosbeta-a cosalpha) *(x-a cosalpha) y-a sinalpha= (sinbeta -sinalpha)(x-a cosalpha)/(cosbeta -cosalpha) y(cosbeta -cosaplha) -asinalpha (cosbeta -cosalpha) = (sinbeta-sinalpha)(x-acosalpha) y(cos beta-cosalpha) =(sinbeta-sinalpha)x- acosalpha sinbeta+asinalpha cosbeta (sinbeta-sinalpha)x +(cosalpha -cosbeta)y +a(sinalpha cosbeta -sinbeta cosalpha)
anonymous
  • anonymous
|dw:1359027554983:dw|Now, Distance between a line Ax+By+C=0 and point (x1,y1) is
anonymous
  • anonymous
Try to use that formula

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
k
anonymous
  • anonymous
A=sinbeta -sinalpha B=cosalpha- cosbeta C=a(sinalpha cosbeta -sinbeta cosalpha) x1=0 y1=0
anonymous
  • anonymous
at the end, i got a(sinalphacosbeta-sinbetacosalpha)/√[2+2cos(alpha+beta)]
anonymous
  • anonymous
asin(alpha-beta)/√[2+2cos(alpha+beta)]

Looking for something else?

Not the answer you are looking for? Search for more explanations.