Given
\qquad \overline{OL}\perp\overline{ON}
OL
⊥
ON
start overline, O, L, end overline, \perp, start overline, O, N, end overline
\qquad m \angle LOM = 3x + 38^\circm∠LOM=3x+38
∘
m, angle, L, O, M, equals, 3, x, plus, 38, degrees
\qquad m \angle MON = 9x + 28^\circm∠MON=9x+28
∘
m, angle, M, O, N, equals, 9, x, plus, 28, degrees
Find m\angle LOMm∠LOMm, angle, L, O, M:
sorry. How about just typing the sides as in AB, CD and ∠ABC without all the repetition and commas ?
and expressions like 9x+28 or 3x+12
all that other junk is just annoying.
You seem to be able to handle ∠, so here's a ⊥ and a ° just to help out
To find m∠LOM, we can use the fact that the sum of the angles in a triangle is 180 degrees. Therefore, we have:
m∠LOM + m∠MON + m∠LON = 180
Substituting the given values, we get:
(3x + 38) + (9x + 28) + 90 = 180
Combine like terms:
12x + 156 = 180
Subtract 156 from both sides:
12x = 24
Divide by 12:
x = 2
Now, substitute the value of x back into m∠LOM:
m∠LOM = 3x + 38
m∠LOM = 3(2) + 38
m∠LOM = 6 + 38
m∠LOM = 44 degrees
Therefore, m∠LOM is equal to 44 degrees.
To find the measure of angle LOM, we need to use the fact that it is a right angle (\(\overline{OL}\perp\overline{ON}\)). Since LOM is a right angle, its measure is 90 degrees.
Therefore, m\angle LOM = 90 degrees.
Given
\qquad \overline{OL}\perp\overline{ON}
OL
⊥
ON
start overline, O, L, end overline, \perp, start overline, O, N, end overline
\qquad m \angle LOM = 3x + 38^\circm∠LOM=3x+38
∘
m, angle, L, O, M, equals, 3, x, plus, 38, degrees
\qquad m \angle MON = 9x + 28^\circm∠MON=9x+28
∘
m, angle, M, O, N, equals, 9, x, plus, 28, degrees
Find m\angle LOMm∠LOMm, angle, L, O, M: