anonymous
  • anonymous
in aright angled triangle, PQR, PQ = 10root3 QR = 20 and p = 90degrees PQT is an equilateral triangle is drawn with base PQ and with vertex T at a maximum possible distance from R, then the length of RT is a. 10root7 b. 10 root 13 c. 10 root 14 d. 10 root 11
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
I think it would be B
anonymous
  • anonymous
could u tell why
anonymous
  • anonymous
|dw:1378823982954:dw| actually the figure is somewhat like this

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phi
  • phi
in 3-D ?
anonymous
  • anonymous
no rt is just line that i connected
anonymous
  • anonymous
it states tha t is at max distance from r
phi
  • phi
ok, in your picture you have PQ = 10, but in the question it says PQ= 10 sqr(3) which is it ?
anonymous
  • anonymous
no pr is 10 pq is 10sqrt(3)
anonymous
  • anonymous
u will find pr to be 10 by pythagorus theorem
phi
  • phi
ok, got it. pq= 10 sqr(3) so all the sides of triangle pqt are 10 sqr(3) the angle Q P T is 90+60= 150 we have 2 sides and an included angle. do you know the law of cosines ?
anonymous
  • anonymous
unfortunately enough i do not but it would really be appreciated if u could solve the ques using whatever law u want i will learn that law later
phi
  • phi
c^2 = a^2 + b^2 - 2 a b cos(C) see http://www.mathsisfun.com/algebra/trig-cosine-law.html
phi
  • phi
can you give it a try ?
phi
  • phi
Here is Khan working an example http://www.khanacademy.org/math/trigonometry/less-basic-trigonometry/law-sines-cosines/v/law-of-cosines-example
anonymous
  • anonymous
i have found the ans thnxzllllllllll
phi
  • phi
you should have gotten \( 10 \sqrt{7} \)
anonymous
  • anonymous
thats what i got

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