Answer:
Step-by-step explanation:
<h2 /><h2>
<u>Consider</u></h2>
<h2>
<u>W</u><u>e</u><u> </u><u>K</u><u>n</u><u>o</u><u>w</u><u>,</u></h2>
So, on substituting all these values, we get
<h2>Hence,</h2>
▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
<h2>ADDITIONAL INFORMATION :-</h2>
Sign of Trigonometric ratios in Quadrants
- sin (90°-θ) = cos θ
- cos (90°-θ) = sin θ
- tan (90°-θ) = cot θ
- csc (90°-θ) = sec θ
- sec (90°-θ) = csc θ
- cot (90°-θ) = tan θ
- sin (90°+θ) = cos θ
- cos (90°+θ) = -sin θ
- tan (90°+θ) = -cot θ
- csc (90°+θ) = sec θ
- sec (90°+θ) = -csc θ
- cot (90°+θ) = -tan θ
- sin (180°-θ) = sin θ
- cos (180°-θ) = -cos θ
- tan (180°-θ) = -tan θ
- csc (180°-θ) = csc θ
- sec (180°-θ) = -sec θ
- cot (180°-θ) = -cot θ
- sin (180°+θ) = -sin θ
- cos (180°+θ) = -cos θ
- tan (180°+θ) = tan θ
- csc (180°+θ) = -csc θ
- sec (180°+θ) = -sec θ
- cot (180°+θ) = cot θ
- sin (270°-θ) = -cos θ
- cos (270°-θ) = -sin θ
- tan (270°-θ) = cot θ
- csc (270°-θ) = -sec θ
- sec (270°-θ) = -csc θ
- cot (270°-θ) = tan θ
- sin (270°+θ) = -cos θ
- cos (270°+θ) = sin θ
- tan (270°+θ) = -cot θ
- csc (270°+θ) = -sec θ
- sec (270°+θ) = cos θ
- cot (270°+θ) = -tan θ
it depends what the number is
Answer: 66 degrees
Explanation:
Check out the attached image below. Figure 1 is the original image without any additions or alterations. Then in figure 2, I extend segment BC to form a line going infinitely in both directions. This line crosses segment DE at point F as shown in the second figure.
Note how angles ABC and DFC are alternate interior angles. Because AB is parallel to DE (given by the arrow markers) this means angle DFC is also 24 degrees
Focus on triangle DFC. This is a right triangle. The 90 degree angle is at C.
So we know that the acute angles x and 24 are complementary. They add to 90. Solve for x
x+24 = 90
x+24-24 = 90-24
x = 66
That is why angle CDE is 66 degrees
Answer:
- (x, y) = (3, 5)
- (x, y) = (1, 2)
Step-by-step explanation:
A nice graphing calculator app makes these trivially simple. (See the first two attachments.) It is available for phones, tablets, and as a web page.
__
The usual methods of solving a system of equations involve <em>elimination</em> or <em>substitution</em>.
There is another method that is relatively easy to use. It is a variation of "Cramer's Rule" and is fully equivalent to <em>elimination</em>. It makes use of a formula applied to the equation coefficients. The pattern of coefficients in the formula, and the formula itself are shown in the third attachment. I like this when the coefficient numbers are "too messy" for elimination or substitution to be used easily. It makes use of the equations in standard form.
_____
1. In standard form, your equations are ...
Then the solution is ...
__
2. In standard form, your equations are ...
Then the solution is ...
_____
<em>Note on Cramer's Rule</em>
The equation you will see for Cramer's Rule applied to a system of 2 equations in 2 unknowns will have the terms in numerator and denominator swapped: ec-bf, for example, instead of bf-ec. This effectively multiplies both numerator and denominator by -1, so has no effect on the result.
The reason for writing the formula in the fashion shown here is that it makes the pattern of multiplications and subtractions easier to remember. Often, you can do the math in your head. This is the method taught by "Vedic maths" and/or "Singapore math." Those teaching methods tend to place more emphasis on mental arithmetic than we do in the US.
The answer is 69.45
hope this helps :)
