I assume that points BC is a straight line and that D is the point where the line DE intersects BC. If this is so, and CDE is 55 degress, then BDE is 125 degrees.
cos(125) = cos(180 - 55) = cos(180)*cos(55) + sin(180)*sin(55) = -cos(55) since cos(180) = -1, and sin(180) = 0 So A = 55 degrees.
-305 degrees would be equal to 55 degrees, which can be found in quadrant one in the standard position.
A triangle with angles measuring 55 degrees, 45 degrees, and 80 degrees is a valid triangle because the sum of its angles equals 180 degrees. This triangle is classified as an acute triangle since all its angles are less than 90 degrees. Additionally, it does not have any sides of equal length, making it a scalene triangle.
30 degrees.
55 degrees.
55 degrees Celsius = 131 degrees Fahrenheit131 degrees F
-55 degrees Celsius is equal to -67 degrees Fahrenheit.
55 degrees Celsius is equal to 131 degrees Fahrenheit.
55 degrees Celsius is equal to 131 degrees Fahrenheit.
55º Fahrenheit = 12,77º Celsius. 1ºCelsius = 33,8º Fahrenheit.
55 degrees Celsius is equal to 131 degrees Fahrenheit.
-55 degrees Celsius = -67 degrees Fahrenheit.
Negative 55 degrees Celsius is equal to -67 degrees Fahrenheit.
55 degrees Celsius is equal to 131 degrees Fahrenheit. This conversion can be calculated using the formula: (°C x 9/5) + 32 = °F.
cos(125) = cos(180 - 55) = cos(180)*cos(55) + sin(180)*sin(55) = -cos(55) since cos(180) = -1, and sin(180) = 0 So A = 55 degrees.
55 degrees Celsius is equal to 131 degrees Fahrenheit.
A parallelogram or a rhombus would fit the given description because they both have equal opposite obtuse angles and equal opposite acute angle that all add up to 360 degrees.