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In the plane figure below, we have a regular octagon $ABCDEFGH$, and I belongs to the diagonal $CG$, so that $\angle GIH = 30°$. Knowing this, determine in degrees the value of the angle indicated by x.(S:$75^o$)

enter image description here

Itry:

$a_i=\frac{180.(8-2)}{8} = 135^o $

$\angle IGH = \frac{135} {2}=67,5^o \implies\angle IHG = 82,5^o$

$\angle AHB = \frac{180-135}{2}=22,5^o \implies \angle BHI =135 - 82,5-22,5 = 30^o \cong \angle GIH$

enter image description here

I'm missing a relationship to finish that I didn't find

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    $\begingroup$ Are you open to a solution using trigonometry? $\endgroup$ Feb 27 at 22:56
  • $\begingroup$ I get 75 degrees, do you know the answer? $\endgroup$ Feb 27 at 23:11
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    $\begingroup$ @RobinSparrow Yes, the correct answer is $75^o$ $\endgroup$ Feb 27 at 23:56
  • $\begingroup$ @DarkLordOfPhysics I would love to see a solution using trigonometry. $\endgroup$
    – Shadow
    Feb 29 at 4:28
  • $\begingroup$ @Shadow I added it below! :) $\endgroup$ Feb 29 at 17:56

2 Answers 2

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Draw $IF$ and $HF$ and by symmetry you have an equilateral triangle so $HI=HF=HB$ so you have an isosceles $\triangle {BHI}$ with 30 degrees in the apex and thus $75$ degrees in the base angles.

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I will present an alternative approach using trigonometry. Let $L$ denote the length of the equal-length sides of the regular octagon $ABCDEFGH$. Applying the Law of Cosines to $\triangle BAH$, we find that $$(HB)^2 = (BA)^2 +(AH)^2 -2(BA)(AH) \cos(\angle BAH)=2L^2(1-\cos(135^{\circ}))$$ $$\implies HB = \left(\sqrt{2+\sqrt{2}}\right) L. \tag{1}$$ Now by the Law of Sines applied to $\triangle GIH$, $$\frac{\sin(\angle GIH)}{HG} = \frac{\sin(\angle HGI)}{IH} \implies \frac{\sin(30^{\circ})}{L} = \frac{\sin(\frac{135^{\circ}}{2})}{IH}$$ $$\implies IH = \left(\sqrt{2+\sqrt{2}}\right) L, \tag{2}$$ where we have made use of the half-angle identity $\sin \left(\frac{\theta}{2}\right)= \pm \sqrt{\frac{1-\cos(\theta)}{2}}$ to compute an exact value for $\sin(67.5^{\circ})$. At this point, we have shown that $HB = \left(\sqrt{2+\sqrt{2}}\right) L = IH$. This is sufficient to conclude that $\angle IBH = \angle HIB$ (formally, we are applying the Law of Sines to $\triangle IBH$). Since the interior angles of $\triangle IBH$ sum to $180^{\circ}$, we moreover have $\angle IBH = \angle HIB = \frac{180^{\circ}-\angle BHI}{2} = 75^{\circ}$.

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  • $\begingroup$ Nice solution! Had been hoping you’d do this after your comment. $\endgroup$ Feb 29 at 21:54

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