Answer:
59.03° is greater than 56.31°' This means that the sliding pole that rises 5 feet for every 3 feet of run making an angle of inclination of 59.03° is steeper than the angle of inclination made by a sliding pole that rises 3 feet for every 2 feet run.
Step-by-step explanation:
Step 1
Express the angle of inclination as follows;
Tan∅=H/R
where;
∅=angle of inclination
H=vertical rise
R=distance of run
Step 2
Determine the angle inclination when;
H=3 feet
R=2 feet
replacing;
Tan∅=(3/2)=1.5
∅=Tan^(-1)(1.5)
∅=56.31°
Determine the angle inclination when;
H=5 feet
R=3 feet
replacing;
Tan∅=(5/3)=1.67
∅=Tan^(-1)(1.67)
∅=59.03°
59.03° is greater than 56.31°' This means that the sliding pole that rises 5 feet for every 3 feet of run making an angle of inclination of 59.03° is steeper than the angle of inclination made by a sliding pole that rises 3 feet for every 2 feet run.
Answer:
I cant see the questions?
Step-by-step explanation:
Answer:
x< -0.24
Step-by-step explanation:
So this is going to sound stupid, but I got x< -0.24 which is just not an option, but this is how I got it...
-25x+14>20
subtract 14 from each side
-25x>6
divide each side by -25
x< -0.24
I switch the way the sign was facing bc I divided by a negative.
Now I know that this must be wrong, but it makes the most sense because I am pretty sure I did all of the math right. Sorry this probably didnt help , but I did a lot of work to try and get the right answer, so Im not putting that work to waste
Answer: option C
Step-by-step explanation:
The diagram of the triangle is shown in the attached photo. The triangle is a right angle triangle ABC
Assuming the given angle is #,
Recalling the trigonometric ratio,
tan # = opposite / adjacent
If tan # = 4, it means
opposite / adjacent = 4/1
Therefore, opposite = 4 and adjacent = 1
Applying Pythagoras theorem,
Hypotenuse^2 = opposite ^2 + adjacent ^2
Hypotenuse = AC
Opposite = 4
Adjacent = 1
AC^2 = 4^2 + 1^2 = 17
AC = ± √17
To determine cos #, we would apply another trigonometric ratio,
Cos# = adjacent /hypotenuse
Cos# = 1/±√17
Cos # =-1 /√17 or 1/√17
