What is the main challenge presented in the video regarding segment DG?

Understanding Secant Ratios in Triangles

Interactive Video
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Mathematics
•
9th - 10th Grade
•
Hard

Amelia Wright
FREE Resource
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10 questions
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1.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
Finding the length without using any trigonometric ratio
Finding the length using the sine ratio
Finding the length without using the cosine ratio
Finding the length using the tangent ratio
2.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
What does the cosine ratio compare in a right triangle?
The hypotenuse to the opposite leg
The adjacent leg to the hypotenuse
The opposite leg to the adjacent leg
The hypotenuse to the adjacent leg
3.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
What is the secant ratio defined as?
The length of the hypotenuse divided by the length of the opposite leg
The length of the hypotenuse divided by the length of the adjacent leg
The length of the adjacent leg divided by the length of the hypotenuse
The length of the opposite leg divided by the length of the adjacent leg
4.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
In which type of triangles can the secant ratio be used?
Only isosceles triangles
Only right triangles
All triangles
Only equilateral triangles
5.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
How is the secant ratio abbreviated in equations?
SNT
SCN
SCT
SEC
6.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
What is the secant of a 35-degree angle approximately equal to in the examples given?
1.00
1.22
1.50
2.00
7.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
What is the first step in solving the original problem using the secant ratio?
Set the sine of 67 degrees equal to x divided by 17
Set the tangent of 67 degrees equal to x divided by 17
Set the cosine of 67 degrees equal to x divided by 17
Set the secant of 67 degrees equal to x divided by 17
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