Question

tell whether the pair of angles are

1. ∠1 and ∠2
2. ∠1 and ∠3
3. ∠1 and ∠4
4. ∠1 and ∠4

find the values of x and y.
25.
26.
∠a and ∠b are complementary

1. m∠a = x°

m∠b=(x - 30)°

Explanation:

Step1: Use vertical - angle property

Vertical angles are equal. So, we set up the equation based on the fact that $16x=(9x + 5)$.
$16x=9x + 5$

Step2: Solve for $x$

Subtract $9x$ from both sides of the equation.
$16x-9x=9x + 5-9x$
$7x=5$
$x=\frac{5}{7}$

Step3: Use linear - pair or vertical - angle relationship for $y$

The sum of angles in a linear - pair is $180^{\circ}$, or we can use the fact that the other pair of vertical angles gives us a relationship. Let's assume the angles $4y$ and one of the other angles are related. Since vertical angles are equal, we first note that the two non - vertical - angle pairs are supplementary. But using vertical angles, we know that the two angles $16x$ and $(9x + 5)$ are vertical, and we can also use the fact that the angle adjacent to $4y$ and one of the given angles forms a linear pair. However, if we assume the other pair of vertical angles, we know that the sum of all angles around the intersection point is $360^{\circ}$. But a simpler way is to assume that the angle adjacent to $4y$ and one of the given angles are supplementary. Since the two given angles $16x$ and $(9x + 5)$ are vertical and equal, we know that the angle adjacent to $4y$ and one of them forms a linear pair. Let's assume the linear - pair relationship. The angle adjacent to $4y$ and $16x$ (or $9x + 5$) gives us:
First, find the value of the angle $16x$ with $x = \frac{5}{7}$, $16\times\frac{5}{7}=\frac{80}{7}$.
The angle adjacent to $4y$ and $\frac{80}{7}$ forms a linear pair. So, $4y+\frac{80}{7}=180$
$4y=180-\frac{80}{7}=\frac{1260 - 80}{7}=\frac{1180}{7}$
$y=\frac{1180}{28}=\frac{295}{7}$

Answer:

$x=\frac{5}{7}$, $y = \frac{295}{7}$