<p>Can anyone help me with number 18 on page 704 and number 15 on page 733 of the blue book? they're triangle problems and I haven't been able to get them for some reason (it is pretty late here). any help/general tips for triangle problems like these would be much appreciated!</p>
<p>Draw a picture and label each of the sides that are congruent. When AB = BC, it is an isosceles triangle meaning that two of the angles will be the same. If angle ABC is 30 degrees, then the other two angles must be (180-30)/2=75. DE=EF=DF means it is an equilateral triangle and therefore, all angles are equal (60 degrees). You are also given that angle BDE is 50 degrees. Angles BDE, EDF, and ADF are supplementary or equal to 180. Therefore, 50+60+x=180. Angle ADF must equal 70. Now angles ADF, DFA, and FAD must equal 180. 70+y+75=180. Y=35. Angle DFA now equals 35, choice B.</p>
<p>thanks so much! if you could help me with the second problem I listed that would be awesome too if you have time</p>
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<li>Supplementary angles add up to 180. Therefore, adjacent angles to n must be 180-n. On the left triangle you now have 180-n for one angle. A triangle has angles adding up to 180 so the other two angles must be 180-(180-n) which equals just n. “n” is equal to two angles. You could probably see that the other triangle would have the same steps so therefore, the degree sum of all 4 angles is 2n.
If you wanted another way to do it, try plugging in your own number. Say the number is 100 for n. Therefore, an adjacent angle would be 80 because of supplementary angles. Going forward, the other two angles would be 180-80 so there it is, 100 degrees in total for 2 angles or 50 degrees for an individual angle. There are 4 angles so 50 times 4 equals 200, which conveniently is 2n. The answer is B.</li>
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