Question 181067
In (triangle)PQR, m(angle)<Q= 90 degrees, PQ=1, and QR=3. The value of sec(angle)<R?
Hey can someone help, dont get it...
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It's a right triangle, so find PR using Pythagoras:
{{{PR = sqrt(1^1 + 3^2)}}}
{{{PR = sqrt(10)}}}
Then, sec(R) = PR/QR
sec(R) = 1/cos(R)
{{{sec(R) = sqrt(10)/3}}}
=~ 1.0541