Question

Can someone please help me with either one of these and show how to solve it? I can’t figure out how to solve any

Answer 1

Answer:

1)  m∠HJG = 130°

2)  m∠XST = 125°

Step-by-step explanation:

Question 1

Given that angles around a point sum to 360°, it follows that:

$m\angle IJH + m\angle HJG + m\angle GJI = 360^{\circ}$

The single arcs at angles ∠IJH and ∠HJG indicate that the two angles have equal measures. So, m∠IJH = m∠HJG.

The measure of a central angle is equal to the measure of the arc it intercepts. Therefore:

$m\angle HJG = (-19x - 3)^{\circ}$

$m\angle IJH = (-19x - 3)^{\circ}$

$m\angle GJI = (x + 107)^{\circ}$

Substitute the angle expressions into the sum equation:

$(-19x - 3)^{\circ} + (-19x - 3)^{\circ} + (x+107)^{\circ} = 360^{\circ}$

Now, solve for x:

$(-19x - 3) + (-19x - 3) + (x+107) = 360\\\\-19x-3-19x-3+x+107=360\\\\101-37x=360\\\\-37x=259\\\\x=-7$

Finally, substitute x = -7 into the expression for angle HJG:

$m\angle HJG=(-19(-7) - 3)^{\circ}\\\\m\angle HJG=(133 - 3)^{\circ}\\\\m\angle HJG=130^{\circ}$

Therefore, the measure of angle HJG is 130°.

$\hrulefill$

Question 2

Line segment TW is the diameter of the given circle.

Given that angles on a straight line sum to 180°, it follows that:

$m\angle XST + m\angle WS\:\!X = 180^{\circ}$

The measure of a central angle is equal to the measure of the arc it intercepts. Therefore:

$m\angle XST = (32x - 3)^{\circ}$

$m\angle WSX = (14x - 1)^{\circ}$

Substitute the angle expressions into the sum equation:

$(32x-3)^{\circ} + (14x-1)^{\circ} =180^{\circ}$

Now, solve for x:

$32x-3+14x-1=180\\\\46x-4=180\\\\46x=184\\\\x=4$

Finally, substitute x = 4 into the expression for angle XST:

$m\angle XST=(32(4) - 3)^{\circ}\\\\m\angle XST=(128- 3)^{\circ}\\\\m\angle XST=125^{\circ}$

Therefore, the measure of angle XST is 125°.