Rays BA and BC are perpendicular. Point D lies in the interior of ∠ABC. If m∠ABD=(3r+5)° and m∠DBC=(5r−27)°, find m∠ABD and m∠DBC.
m∠ABD=
°
m∠DBC=
since AB⊥BC,
3r+5 + 5r-27 = 90
find r, then the angle measures
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To find the measures of angles ABD and DBC, we can use the fact that BA and BC are perpendicular.
Given: m∠ABD = (3r + 5)° and m∠DBC = (5r - 27)°
Since BA and BC are perpendicular, angle ABD and angle DBC are corresponding angles formed by a transversal (BD) intersecting two parallel lines (BA and BC). Therefore, they must be equal.
So, we can set up an equation:
m∠ABD = m∠DBC
(3r + 5)° = (5r - 27)°
Now we can solve this equation to find the value of r.
3r + 5 = 5r - 27
Subtract 3r from both sides:
5 - 3r = -27
Add 3r to both sides:
5 = 2r - 27
Add 27 to both sides:
32 = 2r
Divide both sides by 2:
r = 16
Now that we know the value of r, we can substitute it back into the equations for m∠ABD and m∠DBC:
m∠ABD = (3r + 5)° = (3 * 16 + 5)° = 53°
m∠DBC = (5r - 27)° = (5 * 16 - 27)° = 53°
Therefore,
m∠ABD = 53°
m∠DBC = 53°