## Utilizing Tangents and Chords

We all know that the measure of an

angle inscribed in a circle is half the measure of its intercepted arc.

That is true even when one facet of the angle is tangent to the circle.

Theorem :

If a tangent and a chord

intersect at a degree on a circle, then the measure of every angle fashioned is one

half the measure of its intercepted arc.

Within the diagram, present above, we now have

m∠1 = 1/2 ⋅ m∠arc AB

m∠2 = 1/2 ⋅ m∠arc BCA

## Discovering Angle and Arc Measures

Instance 1 :

Line m is tangent to the circle. Discover the

measure of the crimson angle.

Resolution :

m∠1 = 1/2 ⋅ 150°

m∠1 = 75°

Instance 2 :

Line m is tangent to the circle. Discover the measure of the crimson arc.

Resolution :

m∠arc RSP = 2 ⋅ 130°

m∠arc RSP = 260°

## Discovering an Angle Measure

Instance 3 :

Within the diagram under, BC is

tangent to the circle. Discover m∠CBD.

Resolution :

m∠ CBD = 1/2 ⋅ m∠arc DAB

5x = 1/2 ⋅ (9x + 20)

Multiply both sides by 2.

10x = 9x + 20

Subtract 9x from both sides.

x = 20

So, the angle measure CBD is

m∠ CBD = 5(20°)

m∠ CBD = 100°

## Traces Intersecting Inside or Outdoors a Circle

If two strains intersect a

circle, there are three locations the place the strains can intersect.

We all know discover angle and are measures when

strains intersect on the circle. We will use the next Theorems discover measures when the strains intersect inside or outdoors the circle.

Theorem 1 :

If two chords intersect within the

inside of a circle, then the measure of every angle is one half the sum of the

measures of the arcs intercepted by the angle and its vertical angle.

Within the diagram proven above, we now have

m∠1 = 1/2 ⋅ (m∠arc CD + m∠arc AB)

m∠2 = 1/2 ⋅ (m∠arc BC + m∠arc AD)

Theorem 2 :

If a tangent and a secant, two

tangents, or two secants intersect within the exterior of a circle, then the

measure of the angle fashioned is one half the distinction of the measures of the

intercepted arcs.

Within the diagram proven above, we now have

m∠1 = 1/2 ⋅ (m∠arc BC – m∠arc AC)

Within the diagram proven above, we now have

m∠2 = 1/2 ⋅ (m∠arc PQR – m∠arc PR)

Within the diagram proven above, we now have

m∠3 = 1/2 ⋅ (m∠arc XY – m∠arc WZ)

## Discovering the Measure of an Angle Shaped by Two Chords

Instance 4 :

Discover the worth of x within the diagram proven under.

Resolution :

x° = 1/2 ⋅ (m∠arc PS + m∠arc RQ)

x° = 1/2 ⋅ (106° + 174°)

x° = 1/2 ⋅ (280°)

x° = 140°

Therefore, the worth of x is 140.

## Utilizing Theorem 2

Instance 5 :

Discover the worth of x within the diagram proven under.

Resolution :

Utilizing Theorem 2, we now have

m∠ GHF = 1/2 ⋅ (m∠EDG – m∠GF)

72° = 1/2 ⋅ (200° – x°)

Multiply both sides by 2.

144 = 200 – x

Clear up for x.

x = 56

Instance 6 :

Discover the worth of x within the diagram proven under.

Resolution :

Arcs MN and MLN make an entire circle. So, we now have

m∠arc MLN + m∠MN = 360°

Plug m∠MN = 92°.

m∠arc MLN + 92° = 360°

Subtract 92° from both sides.

m∠arc MLN = 360° – 92°

m∠arc MLN = 268°

Utilizing Theorem 2, we now have

x° = 1/2 ⋅ (m∠MLN – m∠MN)

x° = 1/2 ⋅ (268° – 92°)

x = 1/2 ⋅ (176)

x = 88

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