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

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### 1 Answer

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.**