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Georgia [21]
3 years ago
14

An excellent swimmer is confident he can judge his speed in the water exactly. He bets he can swim two lengths of a 50-meter Oly

mpic size pool at an average speed of precisely 2 meters per second. On his first lane he is a little bit slow and ends up going at 1.5 meters per second. Which average speed does he have to reach on his second lane to win the bet?
Mathematics
1 answer:
Salsk061 [2.6K]3 years ago
6 0
The time needed at 2 meters per second for both rounds would be 100/2 s or 50 seconds. For the first lane he already used 50/1.5 -> 33.33 seconds, so he would need to cover the other 50 meters in exactly 16.67. 50m divided by the 16.67 seconds he has left means he has to swim at a speed of 3 meters per second.
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En un triangulo ABC, el angulo B mide 64° y el angulo C mide 72°. La bisectriz interior CD corta a la altura BH y a la bisectriz
nadezda [96]

Answer:

The difference between the greatest and the smallest angle of the triangle PBQ is 98°.

Step-by-step explanation:

The question is:

In a triangle ABC, angle B measures 64 ° and angle C measures 72°. The inner bisector CD intersects the height BH and the bisector BM at P and Q respectively. Find the difference between the greatest and the smallest angle of the triangle PBQ.

Solution:

Consider the triangle ABC.

The measure of angle A is:

angle A + angle B + angle C = 180°

angle A = 180° - angle B - angle C

             = 180° - 72° - 64°

             = 44°  

It is provided that CD and BM are bisectors.

That:

angle BCP = angle PCH = 36°

angle CBQ = angle QBD = 32°

angle BHC = 90°

Compute the measure of angle HBC as follows:

angle HBC = 180° - angle BHC + angle BCH

                  = 180° - 90° - 72°

                  = 18°

Compute the measure of angle BPC as follows:

angle BPC = 180° - angle PCB + angle CBP

                  = 180° - 18° - 36°

                  = 126°

Then the measure of angle BPQ will be:

angle BPQ = 180° - angle BPC

                  = 180° - 126°

angle BPQ = 54°

Compute the measure of angle PBQ as follows:

angle PBQ = angle B - angle QBD - angle HBC

                  = 64° - 32° - 18°

angle PBQ = 14°

Compute the measure of angle BQP as follows:

angle BQP = 180° - angle PBQ - angle BPQ

                  = 180° - 14° - 54°  

angle BQP = 112°

So, the greatest and the smallest angle of the triangle PBQ are:

angle BQP = 112°

angle PBQ = 14°

Compute the difference:

<em>d</em> = angle BQP - angle PBQ

  = 112° - 14°

  = 98°

Thus, the difference between the greatest and the smallest angle of the triangle PBQ is 98°.

4 0
3 years ago
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