Find cosine of acute angles in a right triangle.

Question

If sides of a right triangle are in Geometric Progression, then find the cosines of acute angles of the triangle.

[Answer] $\frac{\sqrt{5}-1}{2}$,$\sqrt\frac{\sqrt{5}-1}{2}$

My work:

Using Pythagoras Theorem, $a^{2}+a^{2}r^{2}=a^{2}r^{4}$

$1+r^{2}=r^{4}$

$r^{4}-r^{2}-1=0$

Using quadratic equation, $r^{2}=\frac{1\pm\sqrt{1+4}}{2}$

(-) sign rejected because squre cannot be nagetive

so, $r^{2}=\frac{1+\sqrt{5}}{2}$

In right triangle ABC,

$\cos(A)=\frac{ar}{ar^{2}}$=$\frac{1}{r}$=$\frac{1}{\sqrt\frac{1+\sqrt{5}}{2}}$ after rationalization $\sqrt{\frac{2(1-\sqrt{5})}{(1+\sqrt{5})(1-\sqrt{5})}}$ negative in denominator

$\cos(C)=\frac{a}{ar^{2}}$=$\frac{1}{r^{2}}$=$\frac{1}{\frac{1+\sqrt{5}}{2}}$= $\frac{2(1-\sqrt{5})}{(1+\sqrt{5})(1-\sqrt{5})}$=$\frac{2(1-\sqrt{5})}{-4}$= $\frac{\sqrt{5}-1}{2}$

I cannot understand what is wrong with my calculations. Any help is appreciated.

Thanks in advance.

Answer 1

You actually do have the right answers; they just merely look different. A quick check on wolfram alpha will tell you that they are equivalent.

First answer: http://cl.ly/image/1C3F0O3U1A16

Second answer: http://cl.ly/image/0E0X3z3q0M0B

Both of these calculations can be very quickly justified by multiplying both sides of the equation by the denominator and noticing that the conjugates cancel out all of the squareroots.