Question

1. solve for x.
2. solve for x.
3. solve for x.
4. solve for x.
5. if ( mangle fgh = (6x + 21)^circ ) and ( mwidehat{fjh} = (17x - 28)^circ ), find ( mwidehat{fjh} ).
6. if ( mangle stu = (5x - 16)^circ ) and ( mwidehat{su} = (12x - 50)^circ ), find ( mangle stu ).
7. if ( mangle abd = (6x + 26)^circ ) and ( mangle acd = (13x - 9)^circ ), find ( mwidehat{ad} ).
8. if ( mangle kjl = (3x + 2)^circ ) and ( mangle klj = (7x - 32)^circ ), find ( mwidehat{kl} ).

Explanation:

Let's solve problem 13. First, recall the theorem: the measure of an inscribed angle is half the measure of its intercepted arc. Wait, no, in this case, ∠FGH is an inscribed angle? Wait, looking at the diagram (a circle with center K? Wait, no, the diagram shows a circle with points F, G, H, J. Wait, ∠FGH: if G is on the circle, and F, H are on the circle, then ∠FGH is an inscribed angle intercepting arc FJH? Wait, no, maybe it's a tangent or a secant? Wait, the problem says "If \( m\angle FGH=(6x + 21)^\circ \) and \( m\widehat{FJH}=(17x - 28)^\circ \), find \( m\widehat{FJH} \)". Wait, maybe ∠FGH is an inscribed angle, and the arc FJH is the intercepted arc? Wait, no, the measure of an inscribed angle is half the measure of its intercepted arc. Wait, maybe ∠FGH is an inscribed angle, so \( m\angle FGH=\frac{1}{2}m\widehat{FJH} \)? Wait, let's check:

Step 1: Set up the equation

Assuming ∠FGH is an inscribed angle intercepting arc FJH, then:
\( 6x + 21=\frac{1}{2}(17x - 28) \)

Step 2: Solve for x

Multiply both sides by 2:
\( 2(6x + 21)=17x - 28 \)
\( 12x + 42 = 17x - 28 \)
Subtract 12x from both sides:
\( 42 = 5x - 28 \)
Add 28 to both sides:
\( 70 = 5x \)
Divide by 5:
\( x = 14 \)

Step 3: Find \( m\widehat{FJH} \)

Substitute x = 14 into \( 17x - 28 \):
\( 17(14) - 28 = 238 - 28 = 210 \)

Wait, but let's verify. If x=14, then \( m\angle FGH = 6(14)+21 = 84 + 21 = 105^\circ \). Then half of 210 is 105, which matches. So that works.

Answer:

\( 210^\circ \)