Question

find the measure of eb. 25° 96° 118° 146° 7x + 6° 4x + 16°

Explanation:

Step1: Use the inscribed - angle theorem

The measure of an inscribed angle is half the measure of its intercepted arc. The measure of the angle formed by two secants from an external point is half the difference of the measures of the intercepted arcs. Here, $\angle BCD=\frac{1}{2}(\text{arc }BD - \text{arc }BE)$. We know that $\angle BCD = 25^{\circ}$, $\text{arc }BD=7x + 6$ and $\text{arc }BE = 4x+16$.
So, $25=\frac{1}{2}[(7x + 6)-(4x + 16)]$.

Step2: Simplify the right - hand side of the equation

First, simplify the expression inside the brackets: $(7x + 6)-(4x + 16)=7x+6 - 4x-16=3x - 10$. Then our equation becomes $25=\frac{1}{2}(3x - 10)$.
Multiply both sides of the equation by 2 to get rid of the fraction: $2\times25=3x - 10$, which simplifies to $50 = 3x-10$.

Step3: Solve for x

Add 10 to both sides of the equation: $50+10=3x$, so $60 = 3x$. Divide both sides by 3: $x = 20$.

Step4: Find the measure of arc EB

Substitute $x = 20$ into the expression for arc EB. $\text{arc }EB=4x + 16$. So, $\text{arc }EB=4\times20+16=80 + 16=96^{\circ}$.

Answer:

$96^{\circ}$