Question

In AMNO, MO is extended through point O to point P, MNO = (3x + 11)', ZOMN (2x + 20°, and ZNOP = (8x – 5)'. What is the value of x?

Answer 1

The value of x is
`\( (15)/(2) \)`.

Step-by-step explanation:

In the given scenario, we have triangle MNO where MO is extended through point O to point P. Let's denote the angles in the triangle as follows:
`\( \angle MNO = (3x + 11)^\circ \), \( \angle OMN = (2x + 20)^\circ \)`, and
`\( \angle NOP = (8x - 5)^\circ \)`. According to the interior angle sum property of a triangle, the sum of all angles in a triangle is
`\( 180^\circ \)`.

Setting up the equation for the sum of angles in triangle MNO, we get:

(3x + 11) + (2x + 20) + (8x - 5) = 180

Combining like terms, we simplify the equation:

`\[ 13x + 26 = 180 \]`

Solving for x:

`\[ 13x = 154 \]`

`\[ x = (154)/(13) = (22)/(2) = (15)/(2) \]`

Therefore, the value of x is
`\( (15)/(2) \)`. This solution ensures that the sum of interior angles in triangle MNO is
`\( 180^\circ \)`, satisfying the geometric conditions for a triangle.

The solution involves applying basic geometric principles and algebraic manipulation to find the value of x that satisfies the given conditions for the triangle's interior angles. The result
`\( (15)/(2) \)` ensures the geometric consistency of the triangle.