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Filename: Eva Illouz - HARD-CORE romansa.pdf
Solutions to exercise 6.4.2 from strogatz nonlinear dynamics
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7 Hard Trig Problems-Can You Solve These Challenging Gemetry Problems? Updated on June 7, 2016 Trignmry is r f gmry h ls wih lghs gs f ringls. A sc my w-msnl fgrs hr-msnl sl 9
If we use the technique for solvng algebraic equations wit square root and liear terms we get [sin(20)
+
cos(20) - XJA2
=
X A2
si(20)A2 + 2sin(20)cos(20) A = X 2 + 1 1
+
+
1
2cos(20)A2 - 2sin(20)X - cos(20)X
2sn(20)cos(20) - 2sn(20)X - 2cos(20)X X 2 + 1 +
+
+
X A2
X A2
A
2sn(20)cos(20) - 2sin(20)X - 2cos(20) = O X sin(20)cos(20)/[sin(20 =
+
cos(20)] = .250753
Te angles of the green trangle are 90 deges, arctan(X/1)" 14.08 degrees, and arctan(1/X)" 75.92 degees
(2) Find the Missing Angles
Problem: Consider a triangle wth one angle equal to 45 degrees An altitude is drawn to partiton the original trangle into two smaller right triangles, colored yellow and purple n the diagram above If the area of the purple triangle is 15 times the area of the yellow triangle what are the angles of the orginal whole trangle? Solution:
An altitude lne splits the triangle nto two right trangles The yellow trangle s a 45-45-90-degree triangle so its legs have equal length Without any loss of generalit, we can set that lengt equal to 1 If the purple triangle has 15 times the area then its leg lengths are 1 and 15 Call the top unknown angle of the prple triangle x degrees and the bottom unknown angle (90-x) degrees. From trigonometry, we kow that tan(x) x
=
=
151
=
15
arctan(15)" 5631 degrees
Snce 90 - 5631 3369 and 45 + 5631 10131 we now have all the angles of the orignal triangle: 45 33.69 and 101.31 =
=
°
°
(3) Isosceles Integer Triangle
°
si() = (34 I R M th fe g l e = 23 1 •1-.