Question

Find the value of x and then the indicated angle measures.

(CHECK PHOTO ATTACHED‼️‼️)
AND PLEASE PLEASE ANSWER BOTH QUESTIONS‼️

Answer 1

Find the value of x and then the indicated angle measures.

25. If m∠PKR 140, m∠PKT = x + 20, m∠TKR = x - 10, find m∠PKT and m∠TKR?

Answer:

m∠PKT + m∠TKR = m∠PKR

• (x - 10) + (x + 20) = 140

Solving for x,

• x - 10 + x + 20 = 140
• 2x + 10 = 140

subtract 10 from both sides,

• 2x - 10 + 10 = 140 - 10
• 2x = 130

Divide both sides by 2,

• 2x/2 = 130/2
• x = 65

Since,

m∠PKT = x + 20, m∠TKR = x - 10,

Substituting for x, we have

• m∠PKT = 65 + 20
• m∠PKT = 85°
• m∠TKR = 65 - 10
• m∠TKR = 55°

26. If OK bisects ∠POL and ∠POL = 4x-20, m∠POK = x+40, find m∠POL, m∠POK and m∠KOL.

Answer:

Here, OK bisects angle POL.

Angle bisector refers to a line that splits an angle into two equal angles. An bisector of an angle is a ray that divides an angle into two equal parts.

so,

• m∠POK = m∠KOL
• (x + 40) = (x + 40)°

Also,

m∠POK + m∠KOL = m∠POL

• x + 40 + x + 40 = 4x - 20

Solving for x,

• 2x + 80 = 4x - 20

Subtract 2x from both sides,

• 2x - 2x + 80 = 4x - 20 - 2x
• 80 = 2x - 20

add 20 on both sides,

• 80 + 20 = 2x - 20 + 20
• 100 = 2x

Divide both sides by 2,

• x = 50

Since,

• ∠POL = 4x - 20

Substituting for x, we have

• ∠POL = 4x - 20
• ∠POL = 4(50) - 20
• ∠POL = 200 - 20
• ∠POL = 180°

And, m∠KOL = m∠POK

Substituting for x, we have

• m∠KOL = x + 40
• m∠KOL = 50 + 40
• m∠KOL = 90°
• m∠POK = 90°

Answer 2

25. x = 65, m∠TKR = 55°, m∠PKT = 85°

26. x = 50, m∠POL = 180°, m∠POK = 90°, m∠KOL = 90°

Explanation:

25.

We can see from the diagram that ∠PKR is made up of ∠TKR and ∠PKT. The question also tells us that m∠PKR = 140°. Therefore, we can set up the following equation and solve for x:

∠TKR + ∠PKT = ∠PKR

⇒
`(x-10)^\circ + (x+20)^\circ = 140^\circ`

⇒
`x-10 + x+20 = 140` [Removing the brackets]

⇒
`2x+10=140` [Combining like terms]

⇒
`2x = 130` [Subtracting 10 from both sides of the equation]

⇒
`x = (130)/(2)` [Dividing both sides by 2]

⇒
`x = \bf 65`

Now that we have the value of x, we can simply substitute its value into the expressions for ∠TKR and ∠PKT to find their respective values:

m∠TKR = (x - 10)°

= (65 - 10)°

= 55°

m∠PKT = (x + 20)°

= (65 + 20)°

= 85°

26.

The question tells us that the line
`\overrightarrow{OK}` bisects ∠POL. This means the line
`\overrightarrow{OK}` halves the angle POL. This means that ∠POK and ∠KOL are two halves of the angle POL. Therefore, we can set up the following formula and solve for x:

m∠POK =
`(1)/(2)` (∠POL)

⇒
`(x + 40)^\circ = (1)/(2) * (4x - 20)^\circ`

⇒
`x + 40 = 2x - 10`

⇒
`-x + 40 = -10` [Subtracting 2x from both sides]

⇒
`-x = -50`

⇒
`x = \bf 50`

Therefore, we can find the values of ∠POL, ∠POK and ∠KOL using this value x:

m∠POL = (4x - 20)°

= (4 × 50 - 20)°

= (200 - 20)°

= 180°

m∠POK = (x + 40)°

= (50 + 40)°

= 90°

∠POK and ∠KOL are equal as they are halves of ∠POL. Therefore,

m∠KOL = m∠POK = 90°