Rays DA and DC are perpendicular. Point B lies in the interior of
∠
A
D
C
.
If
m
∠
A
D
B
=
(
3
a
+
10
)
°
and
m
∠
B
D
C
=
13
a
°
,
find a,
m
∠
A
D
B
,
and
m
∠
B
D
C
.
S
e
e
h
o
w
a
n
n
o
y
i
n
g
v
e
r
t
i
c
a
l
t
e
x
t
i
s
?
To find the values of a, m∠ADB, and m∠BDC, we can use the fact that the sum of the measures of angles in a triangle is 180°.
Given that rays DA and DC are perpendicular, m∠ADB and m∠BDC are complementary angles, meaning that they add up to 90°.
We are given that m∠ADB = (3a + 10)° and m∠BDC = 13a°.
Since m∠ADB and m∠BDC are complementary, we can write the equation:
(3a + 10)° + 13a° = 90°
Combine like terms:
16a + 10 = 90
Subtract 10 from both sides:
16a = 80
Divide both sides by 16:
a = 5
Now that we have the value of a, we can substitute it back into the given angles to find their measures.
m∠ADB = (3a + 10)°
m∠ADB = (3(5) + 10)°
m∠ADB = (15 + 10)°
m∠ADB = 25°
m∠BDC = 13a°
m∠BDC = 13(5)°
m∠BDC = 65°
Therefore, a = 5, m∠ADB = 25°, and m∠BDC = 65°.
Given that rays DA and DC are perpendicular and point B lies in the interior of ∠ADC, we can solve for the values of a, m∠ADB, and m∠BDC.
Since rays DA and DC are perpendicular, m∠ADB + m∠BDC = 90°.
Given that m∠ADB = (3a + 10)° and m∠BDC = 13a°, we can set up the equation:
(3a + 10)° + 13a° = 90°.
Combining like terms, we have:
16a + 10 = 90.
Subtracting 10 from both sides of the equation, we have:
16a = 80.
Dividing both sides by 16, we have:
a = 5.
Therefore, a = 5.
Now we can substitute the value of a into the expressions for m∠ADB and m∠BDC to find their values.
m∠ADB = (3 × 5 + 10)° = 25°.
m∠BDC = 13 × 5° = 65°.
Therefore, a = 5, m∠ADB = 25°, and m∠BDC = 65°.