Wednesday, June 1, 2016

PARCC Practice Test Question 30 (Day 173)

Question 30 of the PARCC Practice Test is on central and inscribed angles of a circle:

30. Circle T is shown. Line PS and line RS are tangent to circle T.

[The measure of angle PSR is 44 degrees. Point Q is on circle T such that PQR is a major arc -- dw]

Part A:
What is the measure, in degrees, of angle PTR?

Part B:
What is the measure, in degrees, of angle PQR?

For Part A, we notice that three of the angles of quadrilateral PSRT are known -- PSR is given as 44, and SPT and SRT are right angles as tangents are perpendicular to radii. This means that the last angle PTR must be 180-44, or 136 degrees.

For Part B, angle PTR, the central angle has the same measure as arc PR, or 136 degrees, and angle PQR, the inscribed angle, has half that measure, or 68 degrees.

We saw a few questions like this last year. Indeed, we see that there are three types of circle questions that appear on the PARCC -- measurement (as in circumference/area), equation (including completing the square), and angle questions. The angle questions include central angles (taught in Lesson 8-8 of the U of Chicago text), tangent angles (Lesson 13-5), and inscribed angles (Lesson 15-3). This question requires students to be familiar with all three angles.

Notice that the only calculation we performed for Part A was to subtract the given angle, 44 degrees, from 180 degrees. This is because the angles of a quadrilateral add up to 360 degrees and two of the angles are right angles, so the remaining two angles must be supplementary. This idea generalizes, so that if we have two tangent lines that intersect at an angle, the minor arc that the angle subtends always has measure 180 minus the measure of that angle. (And the major arc that the angle subtends always has measure 180 plus the measure of that angle.)

On the other hand, it seems strange that with the PARCC's emphasis on angles, the more advanced circle angle theorems don't appear on the test at all. These are the theorems of Lessons 15-5 and 15-6 of the U of Chicago text -- indeed, we always skip over those lessons on the blog. These theorems are the Angle-Chord Theorem, the Angle-Secant Theorem, the Tangent-Chord Theorem, and finally the Tangent-Secant Theorem. The last of these theorems is:

Tangent-Secant Theorem:
The measure of the angle between two tangents, or between a tangent and a secant, is half the difference of the intercepted arcs.

We've basically already proved the case for two tangents right here in this post. Of the intercepted arcs, one is major and the other is minor. If the angle has measure x, we just explained how the major arc has measure 180 + x and the minor arc has measure 180 - x. The difference between these two measures is 2x, and so half the difference is indeed x, the measure of the angle. In today's problem these two measures are 224 and 136 degrees, and half of this difference is 44 degrees.

PARCC Practice EOY Question 30
U of Chicago Correspondence: Lesson 15-3, The Inscribed Angle Theorem

Key Theorem: Inscribed Angle Theorem

In a circle, the measure of an inscribed arc is one-half the measure of its 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: The U of Chicago text contains some great problems for circles and angles in the SPUR section of Chapter 15, but unfortunately, these questions require theorems from Lessons 15-5 and 15-6, rather than stick to just 13-5 and 15-3. So I came up with my own question -- the diagram actually comes from today's Theoni Pappas calendar question.



2 comments:

  1. Sir, where are the answers so I can compare them to the correct ones?

    ReplyDelete
  2. Here are the answers to the worksheet:

    A. PTR = 136 (explained in my post above)
    B. PQR = 68 (explained in my post above)
    1. DOF = 144
    2. AOD = 153
    3. OBF = 18
    4. DFE = 45
    5. AEOF is a kite. We know that OEA and OFA are right angles (by Radius-Tangent), OE=OF (all radii are =) and OA=OA. Thus Triangles OEA and OFA are congruent by HL. So AE=AF by CPCTC. Combined with OE=OF from above, this is sufficient to make AEOF a kite.
    6. CDOE is a square. The same reasoning as in #5 already tells us that CDOE is at least a kite. Once again, OEC and ODC are right angles by Radius-Tangent. From #5 and CPCTC, OAE=OAF=27, so EAF=54, and since DBF=36, this leaves DCE=90, since the angles of a triangle add up to 180. So far we have three angles of CDOE (OEC, ODC, DCE) are right, thus the fourth is right as well, since the angles of a quad add up to 360. So CDOE is a rectangle -- and it's already a kite as well. A quad that's both a kite and a rectangle must be a square.

    I'm glad that my blog helped you learn Geometry. Please stay healthy during these tough times.

    ReplyDelete