Lesson 7-8: The SAS Inequality (Day 78)

Today on her Mathematics Calendar 2018, Theoni Pappas writes:

Find the pole's height to the nearest meter.

(Here's the given info from the diagram: a wire goes from the top of the pole to a point on the ground 30 m away from the pole. This wire makes an angle 21 degrees 48 minutes with the ground.)

This is an angle of elevation problem similar to the ones which appear in Lesson 14-3 of the U of Chicago text, on the tangent ratio. The only difference is that this problem states the given angle using fractions of a degree. One minute is 1/60 degree. (Degrees, hours, minutes, and seconds all trace back to the Babylonians, who used a sexagesimal, base-60 system.) That's the only trouble that our students will have with this problem (after completing Lesson 14-3).

On the TI-83/84, there are actually ways to enter minutes -- if we press 2nd-MATRX, this shows us the ANGLE menu. The first item is degrees, followed by minutes. To solve this problem, we type:

30tan(21d48m)

using the d (degree) and m (minute) symbols from said ANGLE menu. Or otherwise, we can convert 21d48m to 21.8 degrees:

30tan(21.8)

In either case, put the calculator in degree mode before attempting the tangent. Notice that typing in: 21d48m gives 21.8 directly.

Lesson 7-8 of the U of Chicago text is called "The SAS Inequality." But don't worry -- it doesn't appear in the modern Third Edition of the U of Chicago text at all!

And so this is what I wrote last year about this lesson:

But what is in Section 7-8? This section covers the SAS Inequality -- often known as the Hinge Theorem, according to the Exploration Question at the end of the section. Dr. Franklin Mason also calls this theorem the Hinge Theorem.

I'm of two minds as to whether to include Lesson 7-8 on this blog. Notice that Dr. M includes the Hinge Theorem, Triangle Inequality, and all other inequalities related to triangles in his Chapter 5. But we've been waiting because many of these theorems depend on indirect proof, so we were waiting for Chapter 13. Yet on this blog, I posted the Circumcenter Concurrency Theorem even though it also requires an indirect proof (but, as Dr. Wu pointed out, the part that requires indirect proof can be handwaved over).

In the end, I post Section 7-8 today. If one wants, one can just skip to the review questions.

I wrote more about the significance of the SAS Inequality in my September 11th post. Also, three years ago I included a worksheet with an SSS Inequality -- that's what the "indirect proof" was for. In the end, indirect proof and the SSS Inequality have been dropped. Oh, what the heck -- let me reblog what I wrote about SSS Inequality in this post below, even though it's no longer on the worksheet.

As I mentioned before, this section gives a proof of the SAS Inequality using the Triangle Inequality. As usual, I have decided to convert the proof to two-column format. The figure accompanying the proof gives two triangles, ABC and XYZ.

SAS Inequality Theorem:
If two sides of a triangle are congruent to two sides of a second triangle, and the measure of the included angle of the first triangle is less that the measure of the included angle of the second, then the third side of the first triangle is shorter than the third side of the second.

Given: AB = XY, BC = YZ, angle B < angle Y
Prove: AC < XZ.

Proof:
Statements                                 Reasons
1. AB = XY, etc.                          1. Given
2. exists isometry T s.t. A'B' is   2. Definition of congruent
XY, C' same side of XY as Z
3. C'Y = ZY                                 3. Isometries preserve distance
4. Let m symmetry line C'YZ      4. Isosceles Triangle Symmetry Theorem
(m intersects XZ at Q)       
5. m perp. bis. C'Z                      5. In isosceles triangle, angle bis. = perp. bis.
6. QC' = QZ                                6. Perpendicular Bisector Theorem
7. A'C' < A'Q + QC'                    7. Triangle Inequality
8. AC < XQ + QZ                        8. Substitution
9. XQ + QZ = XZ                        9. Betweenness Theorem (Segment Addition)
10. AC < XZ                                10. Substitution

We see that the proof is similar to that of SAS Congruence Theorem, except that this isometry puts C' on the same side of XY as Z, rather than the opposite side. Dr. M gives two proofs of the SAS Inequality (which he calls "the Hinge Theorem," a name mentioned in Exploration Question 19 in our text) -- his second proof is nearly identical to that given in the U of Chicago. In his first proof, Dr. M uses TASI (Unequal Angles Theorem) directly without invoking the Triangle Inequality -- but we would still be dependent on a theorem not proved until Chapter 13 in the U of Chicago.

Both Dr. M and Glencoe state a converse to SAS Inequality -- Glencoe calls it SSS Inequality. Dr. M hints at this proof -- we can use the same strategy that we used to derive Unequal Angles Theorem from its converse, Unequal Sides Theorem. We prove it indirectly:

SSS Inequality Theorem:
If two sides of a triangle are congruent to two sides of a second triangle, and the third side of the first triangle is shorter than the third side of the second, then the measure of the included angle of the first triangle is less that the measure of the included angle of the second.

Given: AB = XY, BC = YZ, AC < XZ.
Prove: angle B < angle Y

Indirect Proof:
Assume not. Then angle B is either less than or equal to angle Y.

Case 1: angle B = angle Y. Then triangles ABC and XYZ are congruent by SAS Congruence, and so AC = XZ, a contradiction.

Case 2: angle B > angle Y. Then AC > XZ by SAS Inequality, a contradiction.

In either case we have a contradiction of AC < XZ. Therefore angle B < angle Y. QED

For Euclid, the SAS Inequality is his Proposition 24. Dr. M's first proof is based on Euclid:

http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI24.html

and the converse, the SSS Inequality, is Euclid's Proposition 25:

http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI25.html