Question

.) Ray CB bisects angle MCQ. If angle MCB = 10x + 10, and angle BCQ = 20x - 40, find each

X =
Angle MCB =
Angle BCQ =
Angle MCQ =
m
C

Answer 1

Ray CB bisects angle MCQ. With x = 5, Angle MCB is 60°, Angle BCQ is 60°. Angle MCQ is their sum, 120°. The value of x is 5.

If ray CB bisects angle MCQ, it means that angle MCB and angle BCQ are equal since they are divided into two equal parts by the bisector. Therefore, we can set up an equation:

`\[ \text{Angle MCB} = \text{Angle BCQ} \]`

Given that:

`\[ \text{Angle MCB} = 10x + 10 \]`

`\[ \text{Angle BCQ} = 20x - 40 \]`

Setting them equal:

`\[ 10x + 10 = 20x - 40 \]`

Now, solve for x:

`\[ 10x + 10 = 20x - 40 \]`

Subtract 10x from both sides:

`\[ 10 = 10x - 40 \]`

Add 40 to both sides:

`\[ 50 = 10x \]`

Divide by 10:

`\[ x = 5 \]`

Now that we have the value of x, we can find the angles:

1. Angle MCB:

`\[ \text{Angle MCB} = 10x + 10 \]`

`\[ \text{Angle MCB} = 10(5) + 10 \]`

`\[ \text{Angle MCB} = 50 + 10 \]`

`\[ \text{Angle MCB} = 60^\circ \]`

2. Angle BCQ:

`\[ \text{Angle BCQ} = 20x - 40 \]`

`\[ \text{Angle BCQ} = 20(5) - 40 \]`

`\[ \text{Angle BCQ} = 100 - 40 \]`

`\[ \text{Angle BCQ} = 60^\circ \]`

Now, since ray CB bisects angle MCQ, we can find angle MCQ as the sum of angles MCB and BCQ:

`\[ \text{Angle MCQ} = \text{Angle MCB} + \text{Angle BCQ} \]`

`\[ \text{Angle MCQ} = 60^\circ + 60^\circ \]`

`\[ \text{Angle MCQ} = 120^\circ \]`

So, the values are:

`\[ X = 5 \]`

`\[ \text{Angle MCB} = 60^\circ \]`

`\[ \text{Angle BCQ} = 60^\circ \]`

`\[ \text{Angle MCQ} = 120^\circ \]`