Help with these two math problems?
1 answer:
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Answer:
∠NMC = 50°
Step-by-step explanation:
The interpretation of the information given in the question can be seen in the attached images below.
In ΔABC;
∠ A + ∠ B + ∠ C = 180° (sum of angles in a triangle)
∠ A + 70° + 50° = 180°
∠ A = 180° - 70° - 50°
∠ A = 180° - 120°
∠ A = 60°
In ΔAMN ; the base angle are equal , let the base angles be x and y
So; x = y (base angle of an equilateral triangle)
Then;
x + x + 60° = 180°
2x + 60° = 180°
2x = 180° - 60°
2x = 120°
x = 120°/2
x = 60°
∴ x = 60° , y = 60°
In ΔBQC
∠a + ∠e + ∠b = 180°
50° + ∠e + 40° = 180°
∠e = 180° - 50° - 40°
∠e = 180° - 90°
∠e = 90°
At point Q , ∠e = ∠f = ∠g = ∠h = 90° (angles at a point)
∠i = 50° - 40° = 10°
In ΔNQC
∠f + ∠i + ∠j = 180°
90° + 10° + ∠j = 180°
∠j = 180° - 90°-10°
∠j = 180° - 100°
∠j = 80°
From line AC , at point N , ∠y + ∠c + ∠j = 180° (sum of angles on a straight line)
60° + ∠c + ∠80° = 180°
∠c = 180° - 60°-80°
∠c = 180° - 140°
∠c = 40°
Recall that :
At point Q , ∠e = ∠f = ∠g = ∠h = 90° (angles at a point)
Then In Δ NMC ;
∠d + ∠h + ∠c = 180° (sum of angles in a triangle)
∠d + 90° + 40° = 180°
∠d = 180° - 90° -40°
∠d = 180° - 130°
∠d = 50°
Therefore, ∠NMC = ∠d = 50°
There are 4 teams in total and each team has 7 members. One of the team will be the host team.
Tournament committee will be made from 3 members from the host team and 2 members from each of the three remaining teams. Selecting the members for tournament committee is a combinations problem. We have to select 3 members out 7 for host team and 2 members out of 7 from each of the remaining 3 teams.
So total number of possible 9 member tournament committees will be equal to:
This is the case when a host team is fixed. Since any team can be the host team, there are 4 possible ways to select a host team. So the total number of possible 9 member tournament committee will be:
Therefore, there are 2917215 possible 9 member tournament committees