If G is the incenter of triangle ABC, find each measure

If G Is The Incenter Of Triangle ABC, Find Each Measure
If G Is The Incenter Of Triangle ABC, Find Each Measure

Answers

Answer 1

Answer:

m∠ABG = 20 degrees

m∠BCA = 22 degrees

m∠BAC = 118 degrees

m∠BAG = 59 degrees

DG = 4

BE = 12.4

BG = 11.7

GC = 20.4

Step-by-step explanation:

The given parameters are;

m∠CBG = 20°, m∠BCG = 11°

The incenter of a triangle is the point where the three bisectors of ΔABC meets

m∠ABG = m∠CBG = 20° by definition of angle bisector

m∠ABG = 20°

m∠ACG = m∠BCG = 11° by definition of angle bisector

m∠BCA = m∠ACG + m∠BCG = 11° + 11° = 22°

m∠ABC = m∠ABG + m∠CBG = 20° + 20° = 40°

m∠BAC = 180° - (m∠BCA+m∠ABC) = 180° - (40° + 22°) = 118°

m∠BAG = m∠CAG by definition of angle bisector

m∠BAC = 118° = m∠BAG + m∠CAG = m∠BAG + m∠BAG = 2 × m∠BAG

2 × mBAG = 118°

m∠BAG = 118°/2 = 59°

m∠BAG = 59°

Given that "G" is the incenter of the triangle ABC, we have;

GF = GE = DG = The radius of the incircle of the triangle = 4

Therefore, by Pythagoras' theorem, we have;

BG = √(11² + 4²) = √137 ≈ 11.7

BE = √((BG)² + 4²) = √(137 + 4²) = √153 ≈ 12.4

GC = √(20² + 4²) = √416= 4·√26 ≈ 20.4

Answer 2

Applying the definition of the incenter of a triangle, the measures in triangle ABC where G is the incenter are:

[tex]\mathbf{m \angle ABG = 20^{\circ}}[/tex][tex]\mathbf{m \angle BCA = 22^{\circ}}[/tex][tex]\mathbf{m \angle BAC = 118^{\circ}}[/tex][tex]\mathbf{m \angle BAG = 59^{\circ}}[/tex]DG = 4BE = 10.99BG = 11.7GC = 20.4

Recall the following facts about the incenter of a triangle:

The incenter of a triangle is the point where the three angle bisectors that divides each angle vertex into two intersect each other.

The incenter also, is a point equidistant to all three sides of the triangle at right angle.

Applying the properties above, the measures in the image can be calculated as shown below.

Given:

[tex]m \angle EBG = 20^{\circ}\\\\m \angle ECG = 11^{\circ}\\\\CF = 20\\\\EG = 4\\\\BD = 11[/tex]

Find [tex]m \angle ABG[/tex]:

[tex]m \angle ABG = m \angle EBG[/tex] (angle bisector definition)

Substitute

[tex]\mathbf{m \angle ABG = 20^{\circ}}[/tex]

Find [tex]m \angle BCA[/tex]:

[tex]m \angle BCA = 2(m \angle ECG)[/tex] (angle bisector definition)

Substitute

[tex]m \angle BCA = 2(11)\\\\\mathbf{m \angle BCA = 22^{\circ}}[/tex]

Find [tex]m \angle BAC[/tex]:

[tex]m \angle BAC = 180 - (m \angle CBA + m \angle BCA)[/tex] (sum of triangle definition)

Substitute

[tex]m \angle BAC = 180 - (40 + 22)\\\\\mathbf{m \angle BAC = 118^{\circ}}[/tex]

Find [tex]m \angle BAG[/tex]:

[tex]m \angle BAG = \frac{1}{2} (m \angle BAC)[/tex] (angle bisector definition)

Substitute

[tex]m \angle BAG = \frac{1}{2} (118)\\\\\mathbf{m \angle BAG = 59^{\circ}}[/tex]

Find DG:

[tex]DG = EG[/tex] (Perpendicular bisectors of the sides of a the triangle are equal in length from the incenter of a triangle)

Substitute

[tex]\mathbf{DG = 4}[/tex]

Find BG:

[tex]BG = \sqrt{BD^2 + DG^2}[/tex] (Pythagorean Theorem)

Substitute

[tex]BG = \sqrt{11^2 + 4^2} \\\\BG = \sqrt{137} \\\\\mathbf{BG = 11.7}[/tex]

Find BE:

[tex]BE = \sqrt{BG^2 - EG^2}[/tex] (Pythagorean Theorem)

Substitute

[tex]BE = \sqrt{11.7^2 - 4^2} \\\\BE = \sqrt{120.89} \\\\\mathbf{BE = 10.99}[/tex]

Find GC:

[tex]GC = \sqrt{CF^2 + FG^2}[/tex] (Pythagorean Theorem)

Substitute

[tex]GC = \sqrt{20^2 + 4^2} \\\\GC = \sqrt{416} \\\\\mathbf{GC = 20.4}[/tex]

In summary, applying the definition of the incenter of a triangle, the measures in triangle ABC where G is the incenter are:

[tex]\mathbf{m \angle ABG = 20^{\circ}}[/tex][tex]\mathbf{m \angle BCA = 22^{\circ}}[/tex][tex]\mathbf{m \angle BAC = 118^{\circ}}[/tex][tex]\mathbf{m \angle BAG = 59^{\circ}}[/tex]DG = 4BE = 10.99BG = 11.7GC = 20.4

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