Anish: In Holt McDougal Mathematics Grade 7, Common Core Edition
Chapter 4 Proportional Relationships, Chapter Test
Tell whether the figures are similar.
Problem 16. We concluded "No, because the angles disagreed and the sides
weren't proportional."
But a really neat theorem from Trigonometry says that if a triangle has
angles A, B, C, and opposite sides a, b, c, then sin(A)/a = sin(B)/b =
Sin(C)/c.
The first triangle has for these three ratios
In[532]:= N@{Sin[99 º]/11, Sin[27 º]/5, Sin[54 º]/9}
Out[532]= {0.08978984914501252, 0.09079809994790936, 0.08989077715277194}
(You can type º by Esc deg Esc.)
In[533]:= Sin[99 º]/11 == Sin[27 º]/5 == Sin[54 º]/9
Out[533]= False
so it is geometrically impossible! Doubly so, because all three ratios
disagree.
The second triangle has ratios
In[534]:= N@{Sin[102 º]/22, Sin[29 º]/11, Sin[49 º]/17}
Out[534]= {0.04446125457880935, 0.0440736018405761, 0.04439468118957483}
In[535]:= Sin[102 º]/22 == Sin[29 º]/11 == Sin[49 º]/17
Out[535]= False
So the two triangles *are* similar--they're both impossible! Furthermore,
since you
can conclude absolutely anything from a false premise, they are also *not*
similar,
although they're both vegetarian Klingons performing karaoke.
Problem 17. True. Unquestionably. The angles agree. The sides are
proportional.
And the figures are possible because quadrilaterals aren't rigid.
∆WYZ ~ ∆MNO in each pair. Find the unknown measures.
18. <Boggle!> The first triangle is again (doubly) impossible. But by
(alleged)
similarity, we can substitute angles into the second triangle, where the
Law of Sines
gives us two equations and two unknowns, which we can solve!
Out[536]= Sin[37 º]/n == Sin[a]/10 == 1/5 Sin[27 º]
In[537]:= Solve@%
Out[537]=
{{n -> 5 Csc[27 º] Sin[37 º], a -> ConditionalExpression[π - ArcSin[2
Sin[27 º]] + 2 π C[1], C[1] ∈ Integers]},
{n -> 5 Csc[27 º] Sin[37 º], a -> ConditionalExpression[ArcSin[2 Sin[27
º]] + 2 π C[1], C[1] ∈ Integers]}}
Don't be scared by this. It's the best way to write the completely correct
answer.
(Csc[27º] = 1/Sin[27º]. ArcSin[2 Sin[27 º]] is the angle whose sin is
twice the sin of 27º.)
So n -> 5 Csc[27 º] Sin[37 º]
becomes n -> 6.628057453815753, contradicting 5×4/6 = 6.66666... that we got
from proportions in the impossible first triangle.
The ConditionalExpression is Mathematica's slightly awkward notation for two
infinite sets of solutions, where C[1] stands for an arbitrary constant and
C[1] ∈ Integers means it's an integer. By trying C[1] = -1, 0, 1 we would
discover
that our only plausible solution is a ~ 115º, the first case of Out[537]
when C[1] = 0.
Instead of extracting this, we can now use, instead of Solve, the fancier
solver
Reduce, which handles both equations and inequalities. Just tell it a is
obtuse:
In[540]:= Reduce[{90 º < a < 180 º, %536}, {a, n}]
Out[540]= a ==π- ArcSin[2 Sin[27º]] && n == 5 Csc[27º] Sin[37 º]
The π and ArcSin are in radians. In terms of degrees,
In[541]:= % /. π-> 180 º/. Csc@x_ -> 1/Inactive[Sin]@x
Out[541]= a == 180 º - ArcSin[2 Sin[27 º]] && n == (5 Sin[37º])/Sin[27º]
Numerically,
In[544]:= N@Activate@%
Out[544]= a == 2.003152519583483 && n == 6.628057453815753
Again, a is in radians. In degrees,
In[546]:= MapAt[180 Inactive@º#/π &, %%, {1, 2}]
Out[546]= a == 114.7721850931306 º && n == 6.628057453815753
instead of the book's a = 116º and n = 20/3. Nowhere that I can see does
the book admit that these angles and sides are only approximate, nor are
you told how to manipulate approximate quantities.
19. wants you to solve for x and c in the first triangle, which is
"similar" to
the (impossible) second one. By the Law of Sines,
In[524]:= Solve[Sin@c/x == Sin[74 º]/11 == Sin[44 º]/8 == Sin[62 º]/10]
(Debug) Out[524]= {}
Mathematica rejects this as an inconsistent system with no solutions, but
knowing that the second "triangle" is not a triangle, a valid solution is
x = The Square Rutabaga of the Apocalypse, and
c = The Heartbreak of Psoriasis.
--Bill
