## Thursday, April 23, 2015

### PARCC Practice Test Question 8 (Day 153)

Last night I tutored my geometry student. Section 8-6 of the Glencoe text is on the Law of Sines. This topic is not mentioned in the U of Chicago text, and is traditionally a Precalculus topic, yet it's included in the Glencoe text. Of course, we can't help but notice that the Law of Sines is mentioned in the Common Core Standards:

CCSS.MATH.CONTENT.HSG.SRT.D.11
(+) Understand and apply the Law of Sines and the Law of Cosines to find unknown measurements in right and non-right triangles (e.g., surveying problems, resultant forces).

Once again, we have the problem that the Common Core Standards aren't divided into courses. So we don't know whether the Law of Sines is intended to be a Geometry topic or for a later course except by seeing whether there is such a question on the PARCC Geometry Tests. We haven't seen a Law of Sines question yet, but then all of the PARCC questions we've done are non-calculator questions. To remind everyone what the Law of Sines is:

Law of Sines:
In any triangle ABC, the equation a/sin A = b/sin B = c/sin C holds.

My student and I appear to have found an error in one of the Glencoe questions. This is the Glencoe text, not the PARCC exam, but this sounds like the sort of question that could appear on PARCC:

5. SURVEYING. To find the distance between two points A and B that are on opposite sides of a river, a surveyor measures the distance to point C on the same side of the river as point A. The distance from A to C is 240 feet. He then measures the angle across from A to B as 62 degrees and measures the angle across from C to B as 55 degrees. Find the distance from A to B.

Here is the correct answer that my student found:

240/sin 63 = x/sin 55
x = 240 sin 55/sin 63
x = 220.6 ft.

But, as this is an odd-numbered question, here is the answer as given by Glencoe:

240/sin 63 = x/sin 62
x = 240 sin 62/sin 63
x = 237.8 ft.

There are two ways to explain how Glencoe obtained this erroneous answer. The text may have given the length of CB rather than AB. Or the text may have given the length of AB after all, but intended the phrase "angle across from A to B" to refer to angle ACB instead of the natural reading BAC,

One fear Common Core opponents have about the PARCC exam is that there may be poorly worded questions that are ambiguous. Of course, so far the only poorly worded questions I've found have been in the Glencoe text, not on PARCC. But then again, if there is a badly worded question on the PARCC Practice Test we'll never know, because the practice test doesn't provide any answers.

There are a few SSA questions in Glencoe as well as the easier AAS and ASA. Fortunately, all of the SSA questions have the given angle opposite the longer of the two given sides. This is what the U of Chicago calls the SsA case (in the congruence chapter, not the trig chapter). The SsA case never leads to the ambiguous case, so we never have to worry about there being two triangles.

Dr. Franklin Mason also includes a Law of Sines in his text -- in his text that section is numbered 8.7, just one section off from the Glencoe numbering. This implies the possibility that we may be seeing the Law of Sines on the PARCC. Dr. M avoids SSA and therefore the ambiguous case -- yes, the ambiguous case can definitely wait for Precalculus.

But the next question for us to cover is not on the Law of Sines. Question 8 of the PARCC Practice Test is on inscribed angles of a circle. This is the first question of the calculator section of the exam:

The figure shows Triangle ABC inscribed in circle D. If Angle CBD = 44 degrees, find Angle BAC.

Today's question is nowhere near as bad as yesterday's. Since BAC is an inscribed angle, we expect the Inscribed Angle Theorem, included in the U of Chicago's Section 15-3, to be used here. It's because of questions like these that I included 15-3 on the blog -- the only section of the text's final chapter so included.

Notice that CBD is, strictly speaking, an inscribed angle as well. But D is the center of the circle, not a point on the circle, and the point diametrically opposite B on the circle isn't given (so in particular, it's not A). So yes, the arc subtended by CBD would be 88 degrees, but that arc has nothing to do with this problem.

Instead. we must look at Triangle BCD. This looks just like the first part of the proof of the Inscribed Angle Theorem -- the key fact is that Triangle BCD is isosceles, because both DB and DC are radii of the same circle. Then by the Isosceles Triangle Theorem, Angles CBD and BCD are congruent, and so the measure of BCD must also be 44 degrees. Since the measures of the angles of a triangle add up to 180, BDC must measure 92 degrees. This is the central angle that subtends arc BC, which must also measure 92 degrees. Finally, the inscribed angle that intercepts arc BC, angle BAC, must measure half of this, or 46 degrees.

Some people may notice that the given angle is 44 degrees and the correct answer is 46 degrees -- and these two angles add up to 90 degrees. So some people might wonder whether we can prove that Angles CBD and BAC must be complementary. As it turns out, this is the case. Here is the proof of this statement:

Given: Triangle ABC inscribed in circle D
Prove: Angles CBD, BAC are complementary

Statements                              Reasons
1. ABC inscribed in circle D     1. Given
2. DB = DC                            2. Definition of circle
3. Angles CBD = BCD            3. Isosceles Triangle Theorem
4. CBD + BCD + BDC = 180   4. Triangle Angle-Sum Theorem
5. CBD + CBD + BDC = 180   5. Substitution (4 into 5)
6. BDC = 180 - 2m CBD         6. Subtraction Property of Equality
7. Arc BC = Angle BDC          7. Definition of arc measure
8. Angle BAC = 1/2 Arc BC     8. Inscribed Angle Theorem
9. BAC = 1/2 (180 - 2m CBD)  9. Substitution (6 and 7 into 8)
10. BAC = 90 - CBD               10. Distributive Property
11. CBD + BAC = 90              11. Addition Property of Equality
12. CBD, BAC complementary 12. Definition of complementary

This means that all a student has to do is subtract 90 - 44 and enter the number 46 into the test. But it's not at all obvious that angles CBD and BAC should be complementary. The calculation that we performed earlier in this post will be typical for most students.

Unlike most of the problems that we've covered so far this week, traditionalists should have no problem with today's question. Calculating measures of inscribed angles from measures of intercepted arcs is definitely a content-based skill.

PARCC Practice EOY Exam Question 8
U of Chicago Correspondence: Section 15-3, The Inscribed Angle Theorem
Key Theorem: Inscribed Angle Theorem

In a circle, the measure of an inscribed angle is one-half the measure of the intercepted arc.

Common Core Standard:
CCSS.MATH.CONTENT.HSG.C.A.2
Identify and describe relationships among inscribed angles, radii, and chords. Include the relationship between central, inscribed, and circumscribed angles; inscribed angles on a diameter are right angles; the radius of a circle is perpendicular to the tangent where the radius intersects the circle.

Commentary: Most of the inscribed angle problems in the U of Chicago text give the intercepted arc measure, but a few do require students to calculate the arc measure first before finding the inscribed angle measure. In the SPUR section, Objective B, including Questions 7 through 11, are good questions to consider.