If A, B and C are collinear, then
1) if B is between A and C:
AC = AB + BC
AC = 48 + 22 = 70
2) if C is between A and B:
AB = AC + BC
48 = AC + 22 |-22
AC = 26
3) if A is between B and C:
BC = AB + AC
22 = 48 + AC |-48
AC = - 26 < 0 FALSE
Answer:
if B is between A and C, then AC = 70
if C is between A and B, then AC = 26
Answer: The ratio is 2.39, which means that the larger acute angle is 2.39 times the smaller acute angle.
Step-by-step explanation:
I suppose that the "legs" of a triangle rectangle are the cathati.
if L is the length of the shorter leg, 2*L is the length of the longest leg.
Now you can remember the relation:
Tan(a) = (opposite cathetus)/(adjacent cathetus)
Then there is one acute angle calculated as:
Tan(θ) = (shorter leg)/(longer leg)
Tan(φ) = (longer leg)/(shorter leg)
And we want to find the ratio between the measure of the larger acute angle and the smaller acute angle.
Then we need to find θ and φ.
Tan(θ) = L/(2*L)
Tan(θ) = 1/2
θ = Atan(1/2) = 26.57°
Tan(φ) = (2*L)/L
Tan(φ) = 2
φ = Atan(2) = 63.43°
Then the ratio between the larger acute angle and the smaller acute angle is:
R = (63.43°)/(26.57°) = 2.39
This means that the larger acute angle is 2.39 times the smaller acute angle.
Answer:
X=-3
Step-by-step explanation:
5x - 9
-2x + 12
(+) (-)
7x = -21
X= -3
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Answer:
x = 30
Step-by-step explanation:
here 50 is hypotenuse as it is opposite of 90 degree.
x and x + 10 are the two other smaller sides of a right angled triangle respectively.
using pythagoras theorem,
a^2 + b^2 = c^2
x^2 + (x + 10)^2 = 50^2
x^2 + x^2 + 20x + 100 = 2500
2x^2 + 20x + 100 = 2500
2x^2 + 20x + 100 - 2500 = 0
2x^2 + 20x - 2400 = 0
2(x^2 + 10x - 1200) = 0
x^2 + 10x - 1200 = 0
x^2 + (40 - 30) - 1200 = 0
x^2 + 40x - 30x - 1200 = 0
x(x + 40) - 30(x + 40x) = 0
(x + 40)(x - 30) = 0
either x + 40 = 0 OR x - 30 = 0
x = 0 - 40
x = -40
x - 30 = 0
x = 30
x = -40,30
since the length and distance is not measured in negative ur answer will be 30
credit goes to sreedevi102
thank u very much . At first i was wrong and giannathecookie i m really sorry
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