Victoria V. answered • 10/05/17
Tutor
5.0
(402)
Math Teacher: 20 Yrs Teaching/Tutoring CALC 1, PRECALC, ALG 2, TRIG
Hi Ally,
This is a very complicated way to ask for the angles in an isosceles trapezoid!!!!
Do you have any graph paper? Do you have a protractor? You really need to be able to draw this accurately, and if they would let me put a picture here, it would be way easier to see what they want.
Basically, draw a regular hexagon and label the vertices A,E,F,G,H,and L in that order. Draw it well. 120 degrees on each interior angle. Each side must be the same length.
At the side AE, this is where you will attach the regular pentagon ABCE. AE are part of the hexagon, but vertices B,C,and D are all away from the hexagon, and all the interior angles in a regular pentagon are 108 degrees. So draw the pentagons carefully, with all sides the same length and all interior angles 108 degrees.
Do it again at the hexagon side marked HL. HL is a shared side with both the hexagon and a regular pentagon drawn out away from the hexagon. The other vertices to this pentagon will be the I,J, and K.
At the intersection of the two different polygons, at point A, you can make a circle with the 3 angles 108 (from the pentagon), 120 (from the hexagon) and ???? for the open part. 360-108-120 = 132 degrees.
This will also happen at the intersection of the hexagon and the other pentagon at point L. The open part will also be 132 degrees.
NOW... Draw a line from B to K. This will make a trapezoid with 3 congruent sides. This is an isosceles trapezoid, so the two base angles that we do not know (<ABK and <LKB) will each be half of the remaining degrees in any quadrilateral.
Any convex quadrilateral has interior angles that always add up to 360 degrees. ALWAYS. So if we use 132 degrees in each of the obtuse base angles, that leaves 360-132-132 = 96 degrees to split evenly on the two acute base angles.
That makes each of the acute base angles 48 degrees, so m<ABK is 48 degrees.
