AnswersRMA FL-1206310-Geometry Semester 2Special Right Triangles

Week 10: Solving for Side Lengths of Right Triangles Answers

9 verified answers
1
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Which equation could be used to solve for the length of XY?

A
XY = (22)sin(41°)
B
XY = (22)cos(41°)
C
XY =
D
XY =
2
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The measure of angle A is 15°, and the length of side BC is 8. What are the lengths of the other two sides, rounded to the nearest tenth? AC =

A
2.1
B
20.7
C
29.9
D
40.3
3

AB =

A
2.1
B
8.3
C
20.7
D
30.9
4

A 25-foot long ladder is propped against a wall at an angle of 18° with the wall. Which diagram correctly represents this scenario?

A
Option
B
Option
C
Option
D
Option
5

A 25-foot long ladder is propped against a wall at an angle of 18° with the wall. How high up the wall does the ladder reach? Round the answer to the nearest tenth of a foot.

A
13.8 ft
B
23.8 ft
C
26.3 ft
D
80.9 ft
6

Which error did Kevin make?

A
He has the side lengths in the wrong place in the cosine ratio.
B
He multiplied both sides by the length of XZ instead of dividing by XZ.
C
He should have used the sine ratio.
D
He should have used the tangent ratio.
7

Which equations could be used to solve for the unknown lengths of △ABC? Check all that apply.

A
sin(45°) =
B
sin(45°) =
C
9 tan(45°) = AC
D
(AC)sin(45°) = BC
E
cos(45°) =
8

If Diana walks forward and her angle looking to the top of the building changes to 40°, how much closer is she to the building? Round the answer to the nearest tenth of a foot.

A
10.3 ft
B
17.6 ft
C
30.2 ft
D
97.2 ft
9

The building is meters high.

Answer:

55

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