# Find the angle x in the regular octagon below

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:$$75^o$$)

Itry:

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

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

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

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

• Are you open to a solution using trigonometry? Feb 27 at 22:56
• I get 75 degrees, do you know the answer? Feb 27 at 23:11
• @RobinSparrow Yes, the correct answer is $75^o$ Feb 27 at 23:56
• @DarkLordOfPhysics I would love to see a solution using trigonometry. Feb 29 at 4:28
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.
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 $$\angle IBH = \angle HIB = \frac{180^{\circ}-\angle BHI}{2} = 75^{\circ}$$.