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In triangle PQR, sec Q= 2.5, PQ = 3, and QR= 5. Find PR.``

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yahhhh | Student, Grade 12 | Honors

Posted July 12, 2013 at 10:21 PM via web

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In triangle PQR, sec Q= 2.5, PQ = 3, and QR= 5. Find PR.``

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Mary Joy Ripalda | High School Teacher | (Level 3) Educator

Posted July 13, 2013 at 1:42 AM (Answer #1)

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Refer to the figure below which illustrates the given in the problem.

Since we are given with the length of the two sides and its included angle, to solve for PR, apply Cosine Law.

`a^2= b^2+c^2-2abcosA`

Then, let a=PR, b= PQ, c=QP and A=Q.

`(PR)^2= (PR)^2+(PQ)^2-2*(PR)*(QP)cos Q`

`(PR)^2=3^2+5^2-2*3*5*cosQ`

However, the value of angle Q is not given directly. Instead, the given is sec Q=2.5 . And a secant function is reciprocal of cosine.

That means,

`cos Q=1/secQ`

`cosQ=1/2.5`

`cos Q=0.4`

So,

`(PR)^2=3^2+5^2-2*3*5*cosQ`

`(PR)^2=3^2+5^2-2*3*5*0.4`

`(PR)^2=9+25-12`

`(PR)^2=22`

`PR=sqrt22`

Hence, the length of side PR is `sqrt22` units.

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