∠A = 24°
∠B = 87°
∠A = 24°
Explanation:
The sum of angles in a triangle is 180 degrees
let measure of angle A = ∠A
∠B = 15 more than three times the measure of angle A
∠B = 15 + 3∠A
∠C = 45° more than the measure of angle A
∠C = 45° + ∠A
∠A + ∠B + ∠C = 180° (sum of angles in a triangle)
∠A + 15 + 3∠A + 45° + ∠A = 180
collect like terms:
∠A + 3∠A + ∠A + 15 + 45 = 180
5∠A + 60 = 180
5∠A = 180 -60
5∠A = 120
∠A = 120/5
∠A = 24°
∠B = 15 + 3∠A = 15 + 3(24)
∠B = 87°
∠C = 45° + ∠A = 45° + 24°
∠C = 69°
Answer:
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Step-by-step explanation:
abcdefghjklmnopqrstuvwxyz
The measure of an exterior angle of triangle is equal to the sum of measures of two interior angles of triangle that are not supplementary with this exterior angles.
In the case of this question this fact sounds as
m∠XBC = m∠BAC + m∠BCA (option 1).
Now if
- m∠XBC=(3p-6)°
- m∠BAC=(p+4)°
- m∠BCA=84°,
then
3p-6=p+4+84 (option 2),
3p-6=p+88 (option 3),
3p-p-6=p-p+88,
2p-6=88 (option 4),
2p-6+6=88+6,
2p=94 (option 5),
p=47.
Then m∠XBC=(3p-6)°=(3·47-6)°=135°.
Answer: m∠XBC=135°.
The correct answer is 2x^3+10x^2+x+5
I can't understand your question do you have any number there?
