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Geometry Level 4

18 + 6 5 + 10 − 2 5 8 \large \displaystyle\frac{\sqrt{18+6\sqrt{5}}+\sqrt{10-2\sqrt{5}}}{8}

The value of the expression above is equal to the sine of a positive acute angle. What is the measure of this principle angle in degrees?


The answer is 84.

Hobart Pao by Hobart Pao , 23, USA

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1 solution

Mas Mus
Oct 9, 2015

Remember that ( x + y ) + 2 x y = x + y \sqrt{(x+y)+2\sqrt{xy}}=\sqrt{x}+\sqrt{y}

Let

sin ⁡ ( α + β ) = 18 + 6 5 + 10 − 2 5 8 = 1 2 ( 3 + 15 4 ) + 1 2 ( 10 − 2 5 4 ) = 1 2 ( 10 − 2 5 4 ) + 1 2 3 ( 1 + 5 4 ) \begin{array}{c}&\sin(\alpha+\beta)&=\frac{\sqrt{18+6\sqrt{5}}~+~\sqrt{10-2\sqrt{5}}}{8}=\frac{1}{2}\left(\frac{\sqrt{3}+\sqrt{15}}{4}\right)+\frac{1}{2}\left(\frac{\sqrt{10-2\sqrt{5}}}{4}\right)\\&=\frac{1}{2}\left(\frac{\sqrt{10-2\sqrt{5}}}{4}\right)+\frac{1}{2}\sqrt{3}\left(\frac{1+\sqrt{5}}{4}\right)\end{array}

Note that α + β \alpha+\beta is an acute angel so α + β < 9 0 ∘ \alpha+\beta<90^{\circ}

1 2 ( 10 − 2 5 4 ) + 1 2 3 ( 1 + 5 4 ) = sin ⁡ 3 0 ∘ cos ⁡ β + cos ⁡ 3 0 ∘ sin ⁡ β \frac{1}{2}\left(\frac{\sqrt{10-2\sqrt{5}}}{4}\right)+\frac{1}{2}\sqrt{3}\left(\frac{1+\sqrt{5}}{4}\right)=\sin30^{\circ}\cos\beta+\cos30^{\circ}\sin\beta

Now, from this , we know that 1 + 5 4 = sin ⁡ 5 4 ∘ \frac{1+\sqrt{5}}{4}=\sin54^{\circ} . Thus, α + β = 8 4 ∘ \alpha+\beta=84^{\circ}

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