Given:
m∠B = 44°
Let's find the following measures:
m∠A, m∠BCD, m∠CDE
We have:
• m∠A:
Angle A and Angle B are interior angles on same side of a transversal.
The interior angles are supplementary.
Supplementary angles sum up to 180 degrees
Therefore, we have:
m∠A + m∠B = 180
m∠A + 44 = 180
Subtract 44 from both sides:
m∠A + 44 - 44 = 180 - 44
m∠A = 136°
• m,∠,BCD:
m∠BCD = m∠A
Thus, we have:
m∠BCD = 136°
• m∠CDE:
Angle C and angle CDE form a linear pair.
Linear pair of angles are supplementary and supplementary angle sum up to 180 degrees.
Thus, we have:
m∠D = m∠B
m∠D = 44°
m∠CDE + m∠D = 180
m∠CDE + 44 = 180
Subract 44 from both sides:
m∠CDE + 44 - 44 = 180 - 44
m∠CDE = 136°
ANSWER:
• m∠A = 136°
,
•
,
• m∠BCD = 136°
,
•
,
• m∠CDE = 136°
Answer:
48.7%
Step-by-step explanation:
Answer:
81 cm²
Step-by-step explanation:
Since, the lateral face of a triangular pyramid is a triangle,
Given,
The base edge or the base of one lateral face of pyramid, a = 6 cm,
And, the slant height or the height of the face, k = 9 cm,
Thus, the area of one lateral face of the pyramid,
We know that, a Regular triangular pyramid has 3 lateral faces,
Hence, the total lateral area of the pyramid,
3cos (x) / 5 + 4sin (x)/ 5
this is simplified
Multiply exponents together if it is on another exponent
(8^2)^6= 8^12
