Answer:
- 7 is (14x)°
Step-by-step explanation:
In the diagram, the measure of angle 2 is 126°, the measure of angle 4 is (7x)°, and the measure of angle 5 is (4x + 4)°. A transversal intersects 2 lines to form 8 angles. Clockwise from the top left, the angles are 1, 2, 3, 4; 5, 6, 7, 8. What is the measure of angle 7, to the nearest degree? The nearest degree measure of angle 7 is (14x)°
Hello,
Here is the demonstration in the book Person Guide to Mathematic by Khattar Dinesh.
Let's assume
P=cos(a)*cos(2a)*cos(3a)*....*cos(998a)*cos(999a)
Q=sin(a)*sin(2a)*sin(3a)*....*sin(998a)*sin(999a)
As sin x *cos x=sin (2x) /2
P*Q=1/2*sin(2a)*1/2sin(4a)*1/2*sin(6a)*....
*1/2* sin(2*998a)*1/2*sin(2*999a) (there are 999 factors)
= 1/(2^999) * sin(2a)*sin(4a)*...
*sin(998a)*sin(1000a)*sin(1002a)*....*sin(1996a)*sin(1998a)
as sin(x)=-sin(2pi-x) and 2pi=1999a
sin(1000a)=-sin(2pi-1000a)=-sin(1999a-1000a)=-sin(999a)
sin(1002a)=-sin(2pi-1002a)=-sin(1999a-1002a)=-sin(997a)
...
sin(1996a)=-sin(2pi-1996a)=-sin(1999a-1996a)=-sin(3a)
sin(1998a)=-sin(2pi-1998a)=-sin(1999a-1998a)=-sin(a)
So sin(2a)*sin(4a)*...
*sin(998a)*sin(1000a)*sin(1002a)*....*sin(1996a)*sin(1998a)
= sin(a)*sin(2a)*sin(3a)*....*sin(998)*sin(999) since there are 500 sign "-".
Thus
P*Q=1/2^999*Q or Q!=0 then
P=1/(2^999)
The right answer is:
<em>area B = area C</em>
We can solve this problem by using Kepler's laws of planetary motion. There are three Kepler's laws. In this exercise, we need to use the second law. According to this law,<em> a line segment joining a planet and the sun sweeps out equals areas during equals intervals of time. </em>So, a certain planet sweeps out an <em>area B </em>from the point <em>P3 </em>to <em>P4</em> in an interval of time <em>t. </em>On the other hand, for the same interval of time <em>t, </em>the planet sweeps out an <em>area C </em>from point <em>P4</em> to <em>P5, </em>that is equal to the previous area according to second kepler's law.
Answer:
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Step-by-step explanation: