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