2013-2014

Kingspoint Christian School | Trigonometry Final

Study Guide

Trigonometry Final Study Guide Answer Section MULTIPLE CHOICE 1. ANS: A A 42.7% grade means the mountain rises 42.7 ft for every 100 ft of horizontal distance. C

42.7 ft

A

100 ft

B

is the angle the mountain makes with a horizontal line.

Feedback A B C D

Correct! Find the complement of this angle. Convert 42.7% to a fraction. Take the inverse tangent of this fraction. Convert 42.7% to a fraction. Take the inverse tangent of this fraction.

PTS: OBJ: NAT: STA: LOC: KEY: MSC: 2. ANS: 3. ANS:

1 DIF: Average REF: 1bcf0e92-4683-11df-9c7d-001185f0d2ea 8-3.5 Application NT.CCSS.MTH.10.9-12.F.TF.7 | NT.CCSS.MTH.10.9-12.G.SRT.8 WA.WASTD.MTH.08.GEO.G.3.E MTH.C.14.02.03.002 | MTH.C.14.04.03.002 TOP: 8-3 Solving Right Triangles trigonometric ratio | trigonometry | inverse trigonometric function | grade DOK 2 B PTS: 1 A B 2.8 ft

A



C

Write a trigonometric ratio. Substitute the given values. Multiply both sides by AB and divide by sin 15 .

2013-2014

Kingspoint Christian School | Trigonometry Final

feet

Study Guide

Simplify the expression.

Feedback A B C D

Correct! Since the hypotenuse and opposite side are involved, use the sine ratio. The ramp should be substantially larger than the opposite leg, because the opposite leg is across from the smallest possible angle in the triangle. Draw a picture to help determine which trigonometric ratio to use.

PTS: 1 DIF: Average REF: 1bc7e77e-4683-11df-9c7d-001185f0d2ea OBJ: 8-2.5 Problem-Solving Application NAT: NT.CCSS.MTH.10.9-12.G.SRT.8 STA: WA.WASTD.MTH.08.GEO.G.3.E LOC: MTH.C.14.02.03.001 TOP: 8-2 Trigonometric Ratios KEY: trigonometric ratio | trigonometry | side length MSC: DOK 2 4. ANS: D Make sure your calculator is in degree mode. sin 79 = 0.98, cos 47 = 0.68, tan 77 = 4.33 Feedback A B C D

Change your calculator to degree mode. Change your calculator to degree mode. Switch the first and second answers. Correct!

PTS: OBJ: TOP: MSC: 5. ANS:

1 DIF: Basic REF: 1bc349d6-4683-11df-9c7d-001185f0d2ea 8-2.3 Calculating Trigonometric Ratios NAT: NT.CCSS.MTH.10.K-12.5.1 8-2 Trigonometric Ratios KEY: trigonometric ratio | trigonometry | cosine | sine | tangent DOK 2 B 5 is the geometric mean of x and 10. Simplify. y is the geometric mean of the hypotenuse and the segment of the hypotenuse adjacent to y. Simplify. Find the positive square root. z is the geometric mean of the hypotenuse and the segment of the hypotenuse adjacent to z. Simplify. Find the positive square root.

Feedback A B C D

The length of the altitude to the hypotenuse of a right triangle is the geometric mean of the two segments of the hypotenuse. Correct! The leg of a right triangle is the geometric mean of the hypotenuse and the segment of the hypotenuse adjacent to that leg. Create an proportion like this to solve for geometric mean x with regards to parts a and b: x/a = b/x.

2013-2014

PTS: OBJ: TOP: MSC: 6. ANS:

Kingspoint Christian School | Trigonometry Final

Study Guide

1 DIF: Average REF: 1bbbfbb2-4683-11df-9c7d-001185f0d2ea 8-1.3 Finding Side Lengths in Right Triangles LOC: MTH.C.11.08.03.05.006 8-1 Similarity in Right Triangles KEY: similarity | right triangles | altitude to the hypotenuse DOK 2 A Law of Cosines Substitute the given values. Simplify. Take the square root of both sides.

Feedback A B C D

Correct! Take the square root of this answer. Use the Law of Cosines. Change your calculator to degree mode.

PTS: OBJ: NAT: LOC: KEY: 7. ANS:

1 DIF: Average REF: 1bdafa5e-4683-11df-9c7d-001185f0d2ea 8-5.4 Application NT.CCSS.MTH.10.9-12.G.SRT.10 | NT.CCSS.MTH.10.9-12.G.SRT.11 MTH.C.14.06.01.002 TOP: 8-5 Law of Sines and Law of Cosines law of cosines | trigonometry MSC: DOK 2 A Law of Sines Substitute the given values. Cross Products Property Divide both sides by

.

Feedback A B C D

Correct! Multiply 12 by sin 62, then divide by sin 50. Change your calculator to degree mode. Multiply 12 by sin 62, then divide by sin 50.

PTS: 1 DIF: Average REF: 1bd89802-4683-11df-9c7d-001185f0d2ea OBJ: 8-5.2 Using the Law of Sines NAT: NT.CCSS.MTH.10.9-12.G.SRT.10 | NT.CCSS.MTH.10.9-12.G.SRT.11 LOC: MTH.C.14.06.01.001 TOP: 8-5 Law of Sines and Law of Cosines KEY: law of sines | trigonometry MSC: DOK 2 8. ANS: C Use the radii of the circles to find the triangle side lengths. Then use the Law of Cosines to solve for m

.

2013-2014

Kingspoint Christian School | Trigonometry Final

Study Guide

Y X

9 10 11 Z

Feedback A B C D

Use the radii of the circles to find the side lengths of the triangle. Then use the Law of Cosines to find the angle measures. Check the side lengths using the Pythagorean Theorem before deciding the triangle is right. Correct! Use the radii of the circles to find the side lengths of the triangle. Then use the Law of Cosines to find the angle measures.

PTS: NAT: STA: TOP: KEY: 9. ANS:

1 DIF: Advanced REF: 1bdd5cba-4683-11df-9c7d-001185f0d2ea NT.CCSS.MTH.10.9-12.G.SRT.10 | NT.CCSS.MTH.10.9-12.G.SRT.11 WA.WASTD.MTH.08.GEO.G.3.E LOC: MTH.C.14.02.03.002 8-5 Law of Sines and Law of Cosines multi-step | law of cosines | trigonometry MSC: DOK 3 A

By the Pythagorean Theorem,

.

Feedback A B C D

Correct! Use the Pythagorean Theorem to find AB. Sin A = BC/AB. Sin A = BC/AB. Sin A = BC/AB.

PTS: 1 DIF: Average REF: 1bccac36-4683-11df-9c7d-001185f0d2ea OBJ: 8-3.3 Solving Right Triangles STA: WA.WASTD.MTH.08.GEO.G.3.E LOC: MTH.C.14.02.02.002 TOP: 8-3 Solving Right Triangles KEY: trigonometric ratio | trigonometry | solve right triangles MSC: DOK 2 10. ANS: C GH is the length of the hypotenuse of the triangle. You are given FH, which is adjacent to . Since the adjacent side and hypotenuse are involved, use the cosine ratio.

2013-2014

Kingspoint Christian School | Trigonometry Final

Study Guide

Write a trigonometric ratio. Substitute the given values. Multiply both sides by GH and divide by cos 35 . in

Simplify the expression.

Feedback A B C D

The hypotenuse and adjacent side are involved, so a cosine ratio is appropriate. The cosine of an angle is adjacent leg / hypotenuse. Correct! The hypotenuse is involved, so a tangent ratio is not appropriate.

PTS: 1 DIF: Average REF: 1bc58522-4683-11df-9c7d-001185f0d2ea OBJ: 8-2.4 Using Trigonometric Ratios to Find Lengths STA: WA.WASTD.MTH.08.GEO.G.3.E LOC: MTH.C.14.02.03.001 TOP: 8-2 Trigonometric Ratios KEY: trigonometric ratio | trigonometry | side length MSC: DOK 2 11. ANS: C Change the calculator mode to degrees. Feedback A B C D

The first question uses the sine function. Change the calculator mode to degrees. Correct! The second question uses the cosine function.

PTS: 1 DIF: Basic REF: 1bd635a6-4683-11df-9c7d-001185f0d2ea OBJ: 8-5.1 Finding Trigonometric Ratios for Obtuse Angles NAT: NT.CCSS.MTH.10.K-12.5.1 TOP: 8-5 Law of Sines and Law of Cosines KEY: trigonometric ratios | trigonometry MSC: DOK 2 12. ANS: A PTS: 1 13. ANS: A Step 1 Sketch the vectors for the airplane and the wind. N

N

Airplane 400 y 62° Wind

28° x

E

20

E

Step 2 Write the vector for the airplane in component form. The airplane’s vector has a magnitude of 400 mi/h and makes an angle of 28 with the x-axis.

2013-2014

Kingspoint Christian School | Trigonometry Final

, so

.

, so The airplane’s vector is

Study Guide

. .

Step 3 Write the vector for the wind in component form. Since the wind moves 20 mi/h in the negative direction of the x-axis, its vector is

.

Step 4 Find the resultant vector. Add the components of the airplane’s vector and the current’s vector.

The resultant vector in component form is

.

Step 5 Find the magnitude and direction of the resultant vector. The magnitude of the resultant vector is the airplane’s actual speed. mi/h. The angle measure formed by the resultant vector gives the airplane’s actual direction. or N

E

Feedback A B C

D

Correct! When combining the airplane's vector with the wind, subtract 20 from the x-component of the airplane's vector. When combining the airplane's vector with the wind, subtract 20 from the x-component of the airplane's vector. Find the complement of the inverse tangent angle to find the actual direction. Find the component form of the airplane's vector and the wind's vector. Add them together and find the magnitude of the resultant vector. Find the complement of the inverse tangent angle to find the actual direction.

PTS: 1 DIF: Average REF: 1be483ce-4683-11df-9c7d-001185f0d2ea OBJ: 8-6.5 Application NAT: NT.CCSS.MTH.10.9-12.N.VM.2 | NT.CCSS.MTH.10.9-12.N.VM.4.a | NT.CCSS.MTH.10.9-12.N.VM.4.b LOC: MTH.C.10.09.05.01.002 | MTH.C.10.09.05.01.004 TOP: 8-6 Vectors KEY: vector addition | resultant vector MSC: DOK 2 14. ANS: A

2013-2014

R

Kingspoint Christian School | Trigonometry Final

Study Guide

15º

300 ft 15º

S

x

By the Alternate Interior Angles Theorem, m

. From the sketch,

. So

. Feedback A B C D

Correct! Divide 300 by tan 15. Divide 300 by tan 15. Divide 300 by tan 15.

PTS: OBJ: STA: TOP: KEY: 15. ANS:

1 DIF: Average REF: 1bd3d34a-4683-11df-9c7d-001185f0d2ea 8-4.3 Finding Distance by Using Angle of Depression NAT: NT.CCSS.MTH.10.9-12.G.SRT.8 WA.WASTD.MTH.08.GEO.G.3.E LOC: MTH.C.14.02.03.001 8-4 Angles of Elevation and Depression angle of elevation | angle of depression | trigonometry MSC: DOK 2 A

A

23° 

1.8 km  D

23° C

B

Step 1 Draw a sketch. Let B and C represent the beginning and end of the runway. Let CB be the length of the runway. Step 2 Find . By the Alternate Interior Angles Theorem, m In 

,

So

Step 3 Find . By the Alternate Interior Angles Theorem, m In 

,

Step 4 Find

.

So .

So the runway is about 1.2 km long.

.

2013-2014

Kingspoint Christian School | Trigonometry Final

Feedback A B

C

D

Correct! Use right triangles and the tangent function to find the horizontal distance from the pilot to either end of the runway. The length of the runway is the difference between these two lengths. Use right triangles and the tangent function to find the horizontal distance from the pilot to either end of the runway. The length of the runway is the difference between these two lengths. Use right triangles and the tangent function to find the horizontal distance from the pilot to either end of the runway. The length of the runway is the difference between these two lengths.

PTS: 1 DIF: Average REF: 1bd3fa5a-4683-11df-9c7d-001185f0d2ea OBJ: 8-4.4 Application NAT: NT.CCSS.MTH.10.9-12.G.SRT.8 STA: WA.WASTD.MTH.08.GEO.G.3.E LOC: MTH.C.14.02.03.001 TOP: 8-4 Angles of Elevation and Depression KEY: angle of elevation | angle of depression | trigonometry MSC: DOK 2 16. ANS: A Parallel vectors have the same or opposite direction. The vectors

all have the same or opposite direction.

Equal vectors have same direction and the same magnitude. The vectors

are equal.

Feedback A B C D

Correct! Vector HI is also parallel to AB, CD, and JK. FG does not point in the same or opposite direction as AB. Vector JK is also parallel to AB, CD, and HI. Vector HI does not point in the same direction as AB and CD.

PTS: OBJ: LOC: KEY: 17. ANS: 18. ANS:

1 DIF: Average REF: 1be24882-4683-11df-9c7d-001185f0d2ea 8-6.4 Identifying Equal and Parallel Vectors MTH.C.10.09.05.010 | MTH.C.10.09.05.03.009 TOP: 8-6 Vectors parallel vectors | equal vectors MSC: DOK 1 B PTS: 1 A Sine is the ratio of the opposite leg to the hypotenuse. 1.2 is the length of the leg opposite 1.3 is the length of the hypotenuse. 0.5 is the length of the leg adjacent 1.3 is the length of the hypotenuse.

Since

,

2 is

. .

A.

Feedback A B

Correct! The sine of an angle is the length of the opposite leg divided by the length of the

Study Guide

2013-2014

C D

Kingspoint Christian School | Trigonometry Final

Study Guide

hypotenuse. The sine of an angle is the length of the opposite leg divided by the length of the hypotenuse. The sine of an angle is the length of the opposite leg divided by the length of the hypotenuse.

PTS: 1 DIF: Average REF: 1bc80e8e-4683-11df-9c7d-001185f0d2ea OBJ: 8-3.1 Identifying Angles from Trigonometric Ratios STA: WA.WASTD.MTH.08.GEO.G.3.E LOC: MTH.C.14.02.03.002 | MTH.C.14.02.001 TOP: 8-3 Solving Right Triangles KEY: trigonometric ratio | trigonometry MSC: DOK 2 19. ANS: A PTS: 1 20. ANS: C Let x be the height of the radio tower above eye level. 25 is the geometric mean of x and 6. Square 25. Divide both sides by 6. Add 6 to find the height of the tower. The radio tower is about 110 feet tall. Feedback A B C D

The altitude to the hypotenuse of a right triangle is the geometric mean of the two segments of the hypotenuse. Solve for the total height of the tower, not the height above Jody. Correct! Use a proportion with geometric mean to solve.

PTS: OBJ: TOP: MSC: 21. ANS:

1 DIF: Average 8-1.4 Application 8-1 Similarity in Right Triangles DOK 2 A

REF: 1bbe5e0e-4683-11df-9c7d-001185f0d2ea LOC: MTH.C.11.08.03.05.006 KEY: similarity | right triangles | altitude to the hypotenuse

Law of Cosines Substitute the given values. Simplify. Take the square root of both sides.

Feedback A B C D

Correct! Take the square root of this answer. Change your calculator to degree mode. Use the Law of Cosines.

PTS: OBJ: NAT: LOC: KEY:

1 DIF: Average REF: 1bd8bf12-4683-11df-9c7d-001185f0d2ea 8-5.3 Using the Law of Cosines NT.CCSS.MTH.10.9-12.G.SRT.10 | NT.CCSS.MTH.10.9-12.G.SRT.11 MTH.C.14.06.01.002 TOP: 8-5 Law of Sines and Law of Cosines law of cosines | trigonometry MSC: DOK 2

2013-2014

Kingspoint Christian School | Trigonometry Final

22. ANS: A Draw the vector and its components, creating

Study Guide

.

y 3 2

B

1 –4

–3

–2

–1 A –1

1

2

3C 4

x

–2 –3

is the angle formed by the vector and the x-axis, and

. Therefore

Feedback A B

Correct!

C D

Change to degree mode on your calculator.

Find the inverse tangent of

Find the inverse tangent of

PTS: OBJ: TOP: 23. ANS:

.

.

1 DIF: Average REF: 1be22172-4683-11df-9c7d-001185f0d2ea 8-6.3 Finding the Direction of a Vector LOC: MTH.C.14.10.001 8-6 Vectors KEY: vector | direction MSC: DOK 2 C

cos X =

The cosine of an

is

.

Feedback A B C D

Cosine is the ratio of the adjacent side to the hypotenuse. Sine is the ratio of the opposite side to the hypotenuse. Correct! Tangent is the ratio of the opposite to the adjacent side.

PTS: OBJ: STA: LOC: KEY: 24. ANS:

1 DIF: Basic REF: 1bc0c06a-4683-11df-9c7d-001185f0d2ea 8-2.1 Finding Trigonometric Ratios NAT: NT.CCSS.MTH.10.9-12.G.SRT.6 WA.WASTD.MTH.08.GEO.G.3.E MTH.C.14.02.01.002 | MTH.C.14.02.02.004 TOP: 8-2 Trigonometric Ratios trigonometric ratio | trigonometry | cosine MSC: DOK 2 A

.

2013-2014

Kingspoint Christian School | Trigonometry Final

Study Guide

y 3

P

–3

Q

R

3

x

–3

From the figure, QR = 4 and PR = 3.

Feedback A B C D

Correct! Find the complement of this angle. Angle A is the arctangent or inverse tangent of (4/3). Angle A is the arctangent or inverse tangent of (4/3).

PTS: OBJ: LOC: KEY: MSC: 25. ANS:

1 DIF: Average REF: 1bccd346-4683-11df-9c7d-001185f0d2ea 8-3.4 Solving a Right Triangle in the Coordinate Plane MTH.C.14.04.03.001 | MTH.C.14.04.03.002 TOP: 8-3 Solving Right Triangles trigonometric ratio | trigonometry | coordinate geometry | inverse trigonometric ratios DOK 2 A

Feedback A B C D

26. 27. 28. 29.

Correct! Check the definitions of sine and cosine. Check the definitions of sine and cosine. Check the definitions of sine and cosine.

PTS: 1 DIF: Basic REF: 91632d45-6ab2-11e0-9c90-001185f0d2ea OBJ: 8-2-Ext.1 Finding the Sine and Cosine of Acute Angles NAT: NT.CCSS.MTH.10.9-12.G.SRT.7 TOP: 8-2-Ext. Trigonometric Ratios and Complementary Angles KEY: right triangle trigonometry | sine | cosine | tangent MSC: DOK 2 ANS: A PTS: 1 ANS: B PTS: 1 ANS: C PTS: 1 ANS: A Change your calculator to degree mode. Use the inverse trigonometric functions on your calculator to find each angle measure.

Feedback A B C

Correct! Change your calculator to degree mode. Change your calculator to degree mode.

2013-2014

D

Kingspoint Christian School | Trigonometry Final

Study Guide

Switch the first and second answer.

PTS: OBJ: NAT: LOC: TOP: KEY: MSC: 30. ANS:

1 DIF: Basic REF: 1bca49da-4683-11df-9c7d-001185f0d2ea 8-3.2 Calculating Angle Measures from Trigonometric Ratios NT.CCSS.MTH.10.9-12.F.TF.7 MTH.C.14.04.01.002 | MTH.C.14.04.02.002 | MTH.C.14.04.03.002 8-3 Solving Right Triangles trigonometric ratio | trigonometry | inverse trigonometric ratio DOK 1 B

Feedback A B C D

The cosine of an angle is the sine of its complement. Correct! The cosine of an angle is the sine of its complement. The cosine of an angle is the sine of its complement.

PTS: 1 DIF: Average REF: 91635455-6ab2-11e0-9c90-001185f0d2ea OBJ: 8-2-Ext.2 Writing Sine in Cosine Terms and Cosine in Sine Terms NAT: NT.CCSS.MTH.10.9-12.G.SRT.7 TOP: 8-2-Ext. Trigonometric Ratios and Complementary Angles KEY: right triangle trigonometry | cosine | sine MSC: DOK 1 31. ANS: A Draw and label a 30 –60 –90 triangle. The tangent of an angle is

.  2s

60° s

Feedback A B C D

Correct! The tangent of an angle is the ratio opposite leg/ adjacent leg. Use the ratios of the sides of a 30°–60°–90° triangle to create the fraction. The tangent of an angle is the ratio opposite leg/ adjacent leg.

PTS: 1 DIF: Average REF: 1bc322c6-4683-11df-9c7d-001185f0d2ea OBJ: 8-2.2 Finding Trigonometric Ratios in Special Right Triangles NAT: NT.CCSS.MTH.10.9-12.F.TF.3 | NT.CCSS.MTH.10.9-12.G.SRT.6 STA: WA.WASTD.MTH.08.GEO.G.3.E LOC: MTH.C.14.02.02.005 | MTH.C.14.02.02.006 TOP: 8-2 Trigonometric Ratios KEY: trigonometric ratio | trigonometry | tangent | special right triangles | 30-60-90 MSC: DOK 2 32. ANS: A Draw the vector with base at (0, 0) and tip at (6, ). To find the magnitude, use the Distance Formula.

2013-2014

Kingspoint Christian School | Trigonometry Final

Study Guide

Feedback A B C D

Correct! When using the Distance Formula, add the squares of 6 and Take the square root of this answer. Use the Distance Formula with points (0,0) and (6, ).

PTS: OBJ: LOC: KEY: 33. ANS: 34. ANS: 35. ANS:

. The square of

is 9.

1 DIF: Average REF: 1bdfbf16-4683-11df-9c7d-001185f0d2ea 8-6.2 Finding the Magnitude of a Vector MTH.C.10.09.05.03.002 | MTH.C.11.05.04.008 TOP: 8-6 Vectors vector | magnitude MSC: DOK 2 A PTS: 1 B PTS: 1 A B

146.2 m 20º A

x

Use the side opposite the tangent ratio.

and x, and the side adjacent to

Multiply both sides by x and divide both sides by

to write .

Simplify.

Feedback A B C D

Correct! Divide 146.2 by tan 20. Divide 146.2 by tan 20. Change your calculator to degree mode.

PTS: OBJ: STA: TOP: KEY: 36. ANS:

1 DIF: Average REF: 1bd197fe-4683-11df-9c7d-001185f0d2ea 8-4.2 Finding Distance by Using Angle of Elevation NAT: NT.CCSS.MTH.10.9-12.G.SRT.8 WA.WASTD.MTH.08.GEO.G.3.E LOC: MTH.C.14.02.03.001 8-4 Angles of Elevation and Depression angle of elevation | angle of depression | trigonometry MSC: DOK 2 D

Feedback A B C D

Rewrite the equation in terms of either sine or cosine. Then solve. Rewrite the equation in terms of either sine or cosine. Then solve. Rewrite the equation in terms of either sine or cosine. Then solve. Correct!

2013-2014

Kingspoint Christian School | Trigonometry Final

Study Guide

PTS: 1 DIF: Advanced REF: 91658fa0-6ab2-11e0-9c90-001185f0d2ea OBJ: 8-2-Ext.3 Finding Unknown Angles NAT: NT.CCSS.MTH.10.9-12.G.SRT.7 TOP: 8-2-Ext. Trigonometric Ratios and Complementary Angles KEY: trigonometric equation | solve MSC: DOK 3 37. ANS: A PTS: 1 38. ANS: C PTS: 1 39. ANS: A The horizontal change from A to B is 5 units. The vertical change from A to B is –3 units. So the component of

is

.

Feedback A B C D

Correct! The first number is the horizontal component. The horizontal component is 5. The first number is the horizontal component.

PTS: 1 DIF: Average REF: 1bdd83ca-4683-11df-9c7d-001185f0d2ea OBJ: 8-6.1 Writing Vectors in Component Form NAT: NT.CCSS.MTH.10.9-12.N.VM.2 LOC: MTH.C.10.09.05.03.001 TOP: 8-6 Vectors KEY: vector | component form MSC: DOK 2 40. ANS: A 1 and 3 are formed by a horizontal line and a line of sight to a point above the line. They are angles of elevation. 2 and 4 are formed by a horizontal line and a line of sight to a point below the line. They are angles of depression. Feedback A B C D

Correct! An angle of elevation is formed from a horizontal line and a point above the line. An angle of elevation is formed from a horizontal line and a point above the line. An angle of elevation is formed from a horizontal line and a point above the line.

PTS: OBJ: LOC: TOP: KEY:

1 DIF: Basic REF: 1bd170ee-4683-11df-9c7d-001185f0d2ea 8-4.1 Classifying Angles of Elevation and Depression MTH.C.11.02.04.10.001 | MTH.C.11.02.04.10.002 8-4 Angles of Elevation and Depression angle of elevation | angle of depression | trigonometry MSC: DOK 1

NUMERIC RESPONSE 1. ANS: 8.2 PTS: 1 DIF: Advanced LOC: MTH.C.10.09.05.007

REF: 1bebd1f2-4683-11df-9c7d-001185f0d2ea TOP: 8-6 Vectors KEY: vector | magnitude

2013-2014

Kingspoint Christian School | Trigonometry Final

MSC: DOK 2 2. ANS: 12 PTS: NAT: STA: TOP: MSC: 3. ANS: PTS: NAT: LOC: KEY:

1 DIF: Basic REF: 1be94886-4683-11df-9c7d-001185f0d2ea NT.CCSS.MTH.10.9-12.G.SRT.8 | NT.CCSS.MTH.10.9-12.F.TF.7 WA.WASTD.MTH.08.GEO.G.3.E LOC: MTH.C.14.02.03.002 8-3 Solving Right Triangles KEY: inverse tangent | slope | trigonometry DOK 2 324 1 DIF: Advanced REF: 1bebaae2-4683-11df-9c7d-001185f0d2ea NT.CCSS.MTH.10.9-12.G.SRT.8 STA: WA.WASTD.MTH.08.GEO.G.3.E MTH.C.14.02.03.001 TOP: 8-4 Angles of Elevation and Depression angle of elevation | application | slope MSC: DOK 2

Study Guide