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Homework Statement
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
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}<br /> {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 textbook): [itex]\frac{a^2\sqrt{3}}{3}[/itex]
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