An angle measures 73.6° less than the measure of its supplementary angle. What is the measure of each angle?

Mathematics

1 answer:

Vladimir79 [104]3 years ago

6 0

Answer:

Smaller angle = 53.2

Larger angle = 126.8

Step-by-step explanation:

Lets say x is the measure of the supplement. Since we know they're supplementary, we know their angle measure sum will equal 180. We can set up our equation like this $x + (x-73.6) = 180$ . Note: (x - 73.6) is the measure of the smaller angle. By solving, we get 126.8 degrees for the measure of the supplement. If we plug in the value of x into (x-73.6), we get 53.2 degrees as the angle measure of the smaller angle.

You might be interested in

Suppose that five ones and four zeros are arranged around a circle. Between any two equal bits you insert a 0 and between any tw

PolarNik [594]

Answer:

Using backward reasoning we want to show that "We can never get nine 0's".

Step-by-step explanation:

Basically in order to create nine 0's, the previous step had to have all 0's or all 1's. There is no other way possible, because between any two equal bits you insert a 0.

If we consider two cases for the second-to-last step:

There were 9 0's:

We obtain nine 0's if all bits in the previous step were the same, thus all bit were 0's or all bits were 1's. If the previous step contained all 0's, then we have the same case as the current iteration step. Since initially the circle did not contain only 0's, the circle had to contain something else than only 0's at some point and thus there exists a point where the circle contained only 1's.

There were 9 1's:

A circle contains only 1's, if every pair of the consecutive nine digits is different. However this is impossible, because there are five 1's and four 0's (we have an odd number of bits!), thus if the 1's and 0's alternate, then we obtain that 1's that will be next to each other (which would result in a 1 in the next step). Thus, we obtained a contradiction and thus assumption that the circle contains nine 0's after iteratins the procedure is false. This then means that you can never get nine 0's.

To summarize, in order to create nine 0's, the previous step had to have all 0's or al 1's. As we didn't start the arrange with all 0's, the only way is having all 1's, but having all 1's will not be possible in our case since we have an odd number of bits.