1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Expressing the Area of a Triangle in Surd Form

  1. Jul 27, 2012 #1
    1. The problem statement, all variables and given/known data

    Having a bit of trouble with this one, specifically the answer of (ii) (Q. diagram included as attachment). Can anyone help?

    Many thanks.

    Q. In the given diagram, the angles prq & pqs are both right angles. The [itex]\angle[/itex]rpq = X, [itex]\angle[/itex]qps = Y, |pr| = a & |ps| = b. (i) By considering the triangles prq & psq, show that b = [itex]\frac{a}{cosXcosY}[/itex], (ii) If X = 45o & Y = 30o, express the area of [itex]\triangle[/itex]pqs in terms of a. Give your answer in surd form

    3. The attempt at a solution

    Attempt: (i): cos = [itex]\frac{adjacent}{hypotenuse}[/itex]
    Thus, cosX = [itex]\frac{a}{|pq|}[/itex] & cosY = [itex]\frac{|pq|}{b}[/itex]
    Therefore, cosX[itex]\cdot[/itex]cosY = [itex]\frac{a}{|pq|}\cdot\frac{|pq|}{b}[/itex] = [itex]\frac{a}{b}[/itex]
    And so, cosX[itex]\cdot[/itex]cosY = [itex]\frac{a}{b}[/itex] => b = [itex]\frac{a}{cosX\cdot cosY}[/itex]

    (ii) Area of triangle: [itex]\frac{1}{2}b\cdot|pq|\cdot sin30^o[/itex]
    1st, note that: cos45o = [itex]\frac{1}{\sqrt{2}}[/itex] & cos30o = [itex]\frac{\sqrt{3}}{2}[/itex]
    Thus, b = [itex]\frac{a}{cosX\cdot cosY}[/itex] => [itex]\frac{a}
    {1/\sqrt{2}\cdot\sqrt{3}/2}[/itex] => [itex]\frac{a}{\sqrt{3}/2\sqrt{2}}[/itex] => [itex]\frac{2\sqrt{2}a}{\sqrt{3}}[/itex]
    2nd, note that: cosX = [itex]\frac{a}{|pq|}[/itex] => |pq| = [itex]\frac{a}{cos45^o}[/itex] => |pq| = [itex]\frac{a}{1/\sqrt{2}}[/itex] => |pq| = [itex]\sqrt{2}a[/itex]
    Now, [itex]\frac{1}{2}b\cdot|pq|\cdot sin30^o[/itex] = [itex]\frac{1}{2}\cdot\frac{2\sqrt{2}a}{\sqrt{3}}\cdot \sqrt{2}a\cdot\frac{1}{2}[/itex] => [itex]\frac{(2\sqrt{2}a)(\sqrt{2}a)}{4\sqrt{3}}[/itex] => [itex]\frac{(\sqrt{2}a)(\sqrt{2}a)}{2\sqrt{3}}[/itex] => [itex]\frac{2a^2}{2\sqrt{3}}[/itex] => [itex]\frac{a^2}{\sqrt{3}}[/itex]

    Ans.: (From text book): [itex]\frac{a^2\sqrt{3}}{3}[/itex]

    Attached Files:

  2. jcsd
  3. Jul 27, 2012 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I didn't check your steps, but look what happens if you rationalize the denominator in your answer.
  4. Jul 27, 2012 #3
    Great. Thank you very much.
  5. Jul 27, 2012 #4


    Staff: Mentor

    Minor point: You are using => incorrectly in some (not all) places. For example, in the line that starts "Thus, b..." All of the => implications in that line should be =. Likewise in the last line.

    Use = between two expressions that have the same value. Use => between two statements for which the first statement implies the second, such as 2x + 1 = 3 => x = 1.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook