The length of the sides of a right-angled triangle are in geometric progression. What is the ratio of the sines of its acute angles?
- $1$
- $\sqrt {3}$
- $\dfrac{\sqrt{5} + 1}{2}$
- $\sqrt{\dfrac{\sqrt 5 + 1}{2}}$
Let sides be a/r, a, ar. By Pythagoras, (a/r)^2 + a^2 = (ar)^2. Dividing by a^2 gives 1/r^2 + 1 = r^2. Let x = r^2, then 1/x + 1 = x, so x^2 - x - 1 = 0. Solving for x gives r^2 = (1+sqrt(5))/2. The sines of the angles are 1/r and 1/r^2. The ratio is r = sqrt((1+sqrt(5))/2).
Let the sides of the right-angled triangle be a, a over r, and a over r squared, where the largest side a is the hypotenuse and the common ratio is less than one. Applying the Pythagorean theorem gives a squared equals a squared over r squared plus a squared over r to the fourth, and dividing by a squared leaves 1 equals 1 over r squared plus 1 over r to the fourth. Multiplying by r to the fourth and rearranging terms yields the quadratic equation r to the fourth minus r squared minus 1 equals 0, so solving for r squared gives the golden ratio (the square root of 5 plus 1) divided by 2. By the sine rule, the ratio of the sines of the acute angles equals the ratio of their opposite sides, which is the square root of the ratio of the squares of the smaller sides, resulting in the square root of ((the square root of 5 plus 1) divided by 2).