Below are the questions with answers and explanations for Parts 3 and 4 of the **Geometry (Common Core)** Regents exam for January 2016.
Part I questions appeared in a here.
Part II questions appeared in a Here.

### Part III

*Question 32 was answered in the post for Part II. It is reprinted here because it was actually a Part III question.*

**32.** *The aspect ratio (the ratio of screen width to height) of a rectangular flat-screen television I s16:9. The length of the diagonal of the screen is the television’s screen size. Determine and state, to the nearest inch, the screen size (diagonal) of this flat-screen television with a screen height of 20.6 inches. *

This can be solved using ratios and Pythagorean Theorem, or by using Trigonometric Ratios.

First, set up a proportion 16/9 = x/20.6 and cross-multiply.

9x = 329.6, x = 36.6

20.6^{2} + 36.6^{2} = c^{2}

424.36 + 1339.56 = c^{2}

1763.02 = c^{2}

C = 41.999 = 42 inches

Second, the ratio 16:9 represents the opposite over the adjacent, which is the tangent of the top angle of the set. So tan(y) = 16/9 and y = tan-1(16/9) = 60.64 degrees.

Using the adjacent and the hypotenuse and cosine, we get the following:

cos(60.64) = 20.6/x, so x = 9/cos(60.64) = 42.0155… = 42 inches.

**33.** *Given the theorem, "The sum of the the measures of the interior angles of a triangle is 180 ^{o}," complete the proof for this theorem.*

Fill in the missing reasons below.

Fill in the missing reasons below.

This problem caused students a bit of trouble. First, many students had no idea what to write as a reason for Statement (2). What they wrote might have been a correct statement, but not a correct reason for the statement that was made. There is a difference between the two.

Another problem is that many students gave the reason that *"The sum of the angles of a triangle is 180 degrees." * While this is a true statement, it cannot be used as a Reason when that is exactly what you are trying to prove!

Most of the papers I graded score only 2 of the 4 points. (One point for each correct reason. "Given" was preprinted.)

(1) Triangle ABC. Given

(2) Through point C, draw DCE parallel to AB. **Through any line and a point not on the line, there is exactly one line passing through that point parallel to the line.**

(3)m<l = m<ACD, m<3 = m<BCE **When two parallel lines are cut by a transversal, alternate interior angles are congruent.**

(4) m<ACD + m<2 + m<BCE = 180° **The sum of angles creating a straight line is 180 degrees.**

(5) mLl + mL2 + mL3 = 180° **Substitution Property**

I saw many students who didn't label angles as "alternate interior", leaving out one word or the other. And a number of them mentioned *supplementary angles* in statement (4), but "supplementary" applies to **two** angles, not three.

**34.** *Triangle XYZ is shown below. Using a compass and straightedge, on the line below, construct and label triangle ABC, such that triangle ABC = triangle XYZ. [Leave all construction marks.]
Based on your construction, state the theorem that justifies why triangle ABC is congruent to triangle XYZ. *

Sorry, but I can't illustrate constructions very well. I promise to get back to it.

Simple steps: From point X, measure the distance to Y. On the line, make point A and mark the distance to with an arc, and label where the arc intersects the line as B. Go back to X and measure the distance to Z. Go to X and make another arc. Go back to Y and measure the length to Z. Put the compass on B and make an arc. Where the two arcs intercept, label that point C. Using the straightedge, make lines AC and BC. The theorem used was SSS.

Note: It is possible that you could have used SAS to create the triangles. Whichever theorem you stated, the construction marks had to match it to get full credit.

You also could have gotten 1 point for saying SSS, SAS, or ASA without any construction at all.

### Part IV

**35.** *Given: Parallelogram ANDR with AW and DE bisecting NWD and REA at points W and E, respectively*

Prove that triangle ANW = triangle DRE.

Prove that quadrilateral AWDE is a parallelogram.

Prove that triangle ANW = triangle DRE.

Prove that quadrilateral AWDE is a parallelogram.

There are many variations of this proof which will be acceptable. When using SAS to prove two triangles congruent, be sure that you have included statements regarding the two pairs of sides and the included pairs of angles.

(1) Parallelogram ANDR with AW and DE bisecting NWD and REA. **Given**

(2) AE = RE and DW = NW **Definition of segment bisector.**

(3) RA = DN and RD = NA **Opposite sides of a parallelogram are congruent.**

(4) RE = NW **Halves of congruent segments are congruent.**

(5) Angle R = Angle N **Opposite angles of a parallelogram are congruent.**

(6) Triangle DRE = Triangle ANW ** SAS**

(7)ED = WA **CPCTC**

(8)AE = DW **Halves of congruent segments are congruent.**

(9) AWDE is a parallelogram **If both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is a parallelogram**

Parallelograms have many properties that apply only to them. Proving any of those would be sufficient proof that AWDE was a parallelogram.

**36.** *Cathy wants to determine the height of the flagpole shown in the diagram below. She uses a survey instrument to measure the angle of elevation to the top of the flagpole, and determines it to be 34.9°. She walks 8 meters closer and determines the new measure of the angle of elevation to be 52.8°. At each measurement, the survey instrument is 1.7 meters above the ground.*

Determine and state, to the nearest tenth of a meter, the height of the flagpole.

Determine and state, to the nearest tenth of a meter, the height of the flagpole.

Personally, I hate problems that force you to add 1.7 meters at the end for no good reason. It's just an added step for people to forget about and lose a point.

One also wonder why Cathy didn't walk **to** the pole and measure that distance rather than walking an extra 8 meters **away** from it. But this is the problem we are given.

We are looking for the height of the pole. That is the opposite side to both angles. We have some information about the distance along the ground. That is the adjacent side to the angles. We have no information about the hypotenuse, nor are we looking for it.

Opposite and adjacent means that we are using **tangent**.

Let h be the height and x be the distance along the ground to the first measurement.

tan 52.8 = h / x

so h = x (tan 52.8)

And tan 34.9 = h / (x + 8)

so h = (x + 8)(tan 34.9)

This means that x (tan 52.8) = (x + 8)(tan 34.9). We need to isolate x.

Distribute: x (tan 52.8) = x (tan 34.9) + 8(tan 34.9)

Subtract: x (tan 52.8) - x (tan 34.9) = 8(tan 34.9)

Factor: x (tan 52.8 - tan 34.9) = 8(tan 34.9)

Divide: x = 8(tan 34.9) / (tan 52.8 - tan 34.9)

Calculate: x = 9.00371 = 9 meters

Use x to get h:

h = 9.00371(tan 52.8) = 11.8619

And the height of the survey instrument: 11.86 + 1.7 = 13.56 = **13.6 meters**.

An amazing amount of work, but it was worth 6 points.

You lost credit if you forgot the 1.7 meters, used the wrong functions, or rounded in the middle of the problem so that your answer didn't round to 13.6

Oddly, if you multiplied 1.7 * 8 you get *EXACTLY* 13.6. If you wrote 13.6 with NO WORK WHATSOEVER, you got one point. If you wrote 1.7 * 8 = 13.6, you got ZERO POINTS for a totally incorrect response or a correct response found through a totally incorrect method. Those are the breaks.

**END OF PART IV.
**

How did you do?

## 6 comments:

On question 36,I think using Law of Sines was easier then the method of using two equations involving tangent. I know Law of sines is optional topic but I teach it to my students and after this question I'm happy I do.

To be perfectly honest, because I haven't checked the curriculum lately.

I haven't taught a full year of CC Geometry since it has been implemented and I was unaware that Law of Sines was part of it now.

Yes, that would have made the solution easier.

That said, this solution works for anyone who forget it, or their teachers didn't cover it enough.

I'll add that when I get the chance.

How did you 'Divide: x = 8(tan 34.9) / (tan 52.8 - tan 34.9) ' for question 36?

I'm kinda stuck at that part.

I'm not sure I understand your question.

In the previous step, we factored and got x times (tan 52.8 - tan 34.9)

To isolate the x (solve for x), we divide both sides by (tan 52.8 - tan 34.9)

To do the calculation, put the entire thing, with the parentheses, into your graphing calculator. If you're getting a mistake, check that you are in degree mode and not radian more.

You need to type 8*(tan(34.9))/(tan(52.8)-tan(34.9))

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