PROBLEMS INVOLVING VERTICALLY OPPOSITE ANGLES

Find the value(s) of the variable(s) :

Example 1 :

Solution :

In the given figure,

<APC  =  <BPD (vertically opposite angles)

Here <APC  =  105˚ and <BPD  =  (2x – 11)˚ 

105˚  =  (2x – 11)˚

105˚ + 11˚  =  2x˚

116˚  =  2x

x  =  58˚

So, the value of x is 58˚

Example 2 :

Solution  :

In the given figure,

<APD  =  (6x + 19)˚ (vertical angle)

<BPD  =  x˚ (adjacent angle)

We know that a vertical angle and its adjacent angle are supplementary angles.

So,

<APD + <BPD  =  180˚

 (6x + 19)˚ + x˚  =  180˚

6x˚ + 19˚ + x˚  =  180˚

7x˚  =  180˚ - 19˚

7x˚  =  161˚

x  =  23˚

So, the value of x is 23˚

Example 3 :

Solution  :

In the given figure,

<APC  =  <BPD (vertically opposite angles)

Here <APC  =  78˚ and <BPD  =  (5x – 2)˚ 

78˚  =  (5x – 2)˚

78˚ + 2˚  =  5x˚

80˚  =  5x

x  =  16˚

So, the value of x is 16˚

Example 4  :

Solution :

In the given figure,

<APD  =  (6x - 32)˚ (vertical angle)

<BPD  =  (2x – 20)˚ (adjacent angle)

We know that a vertical angle and its adjacent angle are supplementary angles.

So,

<APD + <BPD  =  180˚

6x - 32˚+ 2x - 20˚  =  180˚

8x – 52˚  =  180˚

8x  =  232˚

x  =  29˚

Then,

<APC  =  (y – 12)˚ (vertical angle)

<BPC  =  (3y – 8)˚ (adjacent angle)

So,

<APC + <BPC  =  180˚

y – 12˚ + 3y – 8˚  =  180˚

4y - 20˚  =  180˚

4y  =  180˚ + 20˚

4y  =  200˚

y  =  50˚

Example 5 :

Solution  :

In the given figure,

<APD  =  (4x + 10)˚ (vertical angle)

<BPD  =  (3x – 5)˚ (adjacent angle)

So,

<APD + <BPD  =  180˚

4x + 10˚+ 3x - 5˚  =  180˚

7x + 5˚  =  180˚

7x  =  180˚ - 5˚

7x  =  175˚

x  =  25˚

Then,

<APC  =  (2y + 28)˚ (vertical angle)

<BPC  =  (4y + 26)˚ (adjacent angle)

<APC + <BPC  =  180˚

2y + 28˚ + 4y + 26˚  =  180˚

6y + 54˚  =  180˚

6y  =  180˚ - 54˚

6y  =  126˚

y  =  21˚

Example 6 :

Solution :

In the given figure,

<APD  =  (7x - 248)˚ (vertical angle)

<BPD  =  (x + 44)˚ (adjacent angle)

So,

<APD + <BPD  =  180˚

7x - 248˚+ x + 44˚  =  180˚

8x - 204˚  =  180˚

8x  =  180˚ + 204˚

8x  =  384˚

x  =  48˚

Then,

<APC  =  (9y - 187)˚ (vertical angle)

<BPC  =  (11y - 253)˚ (adjacent angle)

So,

<APC + <BPC  =  180˚

9y - 187˚ + 11y - 253˚  =  180˚

20y - 440˚  =  180˚

20y  =  180˚ + 440˚

20y  =  620˚

y  =  31˚

Example 7 :

Solution :

In the given figure,

<APD  =  <BPC (vertically opposite angles)

Here <APD  =  (3x + 20)˚ and <BPD  =  (5x – 50)˚ 

3x + 20˚  =  5x – 50˚

3x - 5x  =  -50˚ - 20˚

-2x  =  -70˚

x  =  35˚

Then,

<APC  =  y˚ (vertical angle)

<BPC  =  (5x - 50)˚ (adjacent angle)

So,

<APC + <BPC  =  180˚

y˚ + 5x - 50˚  =  180˚

y˚ + 5(35˚) - 50˚  =  180˚

y + 175˚ - 50˚  =  180˚

y + 125˚  =  180˚

y  =  180˚ - 125˚

y  =  55˚

Example 8 :

Solution :

In the given figure,

<APC  =  <BPD (vertically opposite angles)

Here <APC  =  6x˚ and <BPD  =  (4x + 16)˚ 

6x˚  =  4x + 16˚

6x - 4x  =  16˚

2x  =  16˚

x  =  8˚

Then,

<APC  =  6x˚ (vertical angle)

<BPC  =  11y˚ (adjacent angle)

So,

<APC + <BPC  =  180˚

6x˚ + 11y˚  =  180˚

6(8˚) + 11y˚  =  180˚

48˚ + 11y˚  =  180˚

11y˚  =  180˚ - 48˚

11y  =  132˚

y  =  12˚

Example 9 :

Solution :

In the given figure,

<APC  =  <BPD (vertically opposite angles)

Here <APC  =  7x˚ and <BPD  =  56˚ 

7x˚  =  56˚

x  =  8˚

Then,

<CPE  =  2x˚ (vertical angle)

<BPE  =  y˚ (adjacent angle)

So,

<APC + <BPC  =  180˚

2x˚ + y˚  =  180˚

2(8˚) + y˚  =  180˚

16˚ + y˚  =  180˚

y˚  =  180˚ - 16˚

y  =  164˚

Example 10 :

l and m are parallel and a line t intersects these lines at P and Q, respectively. Find the sum 2a + b.

Solution :

• 132 and b are vertically opposite angles.
• 132 and a are corresponding angles.

Example 11 :

In the accompanying diagram, parallel lines 𝐻𝐸 and 𝐴𝐷 are intersected by transversal 𝐵𝐹 at G and C, respectively.

If 𝑚 < 𝐻𝐺𝐹 = 5𝑛 and 𝑚 < 𝐵𝐶𝐷 = 2𝑛 + 66, what is 𝑚 < 𝐻𝐺𝐹 and 𝑚 < 𝐹𝐺𝐸?

Solution :

<HGF = <ACG (vertically opposite angles)

5n = 2n + 66

5n - 2n = 66

3n = 66

n = 66/3

n = 22

𝑚 < 𝐻𝐺𝐹 = 5(22)

= 110

𝑚 < 𝐹𝐺𝐸 = 180 - 110

= 70

Example 12 :

Given: 𝑚 < 3 = 3𝑥 + 30 and 𝑚 < 7 = 5𝑥

a) The relationship is:

b) The equation is:

c) Angle 3

Solution :

𝑚 < 3 = 3𝑥 + 30 and 𝑚 < 7 = 5𝑥

These angles are corresponding angles.

3x + 30 = 5x 

30 = 5x - 3x

30 = 2x

x = 30/2

x = 15

To find angle measure 3, we apply the value of x in 3x + 30

= 3(15) + 30

 = 45 + 30

= 75

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