84/21 = 4 20 * 4 = 80 So the base edge of the second pyramid is 80. Hope this helps :)
Answer:
<h2>D.
x ≈ 31.3°</h2>
Step-by-step explanation:
The triangle CBV is angled triangle with right angle at ∠VBC.
Since ∠VBC = (3x-4) °
To get the value of x, we will equate the angle ∠VBC to 90°
This results into (3x-4) ° = 90
Simplifying the resulting equation;
3x-4 = 90
Adding 4 to both sides;
3x-4+4 = 90+4
3x = 94
Dividing both sides by 3
3x/3 = 94/3
x = 31.33°
x ≈ 31.3°
We know angle P is 90 degrees because of Thales Theorem, and angle N is 60 degrees. Because the angles in a triangle have to add up to 180 degrees, angle PMN has to be 30 degrees.
Angle QMO is half of angle PMN, so it's 15 degrees. Angle MQO is equal to angle QMO because the radii make Triangle OQM isosceles, so angle MQO is 15 degrees.
I believe that the answer is p+7
