use your notes on incenters to complete question b10-b17. directions: if g is the incenter of δabc, find each measure. 10. ( mangle abg ) 11. ( mangle bca ) 12. ( mangle bac ) 13. ( mangle bag ) 14. ( dg ) 15. ( be ) 16. ( bg ) 17. ( gc ) diagram of triangle abc with incenter g, right angles at d, e, f, lengths 11, 4, 20, angles 20°, etc.

Question

Accepted Answer

To solve these problems, we use the properties of the incenter of a triangle. The incenter is the intersection of the angle bisectors, so it is equidistant from all sides, and it bisects the angles of the triangle. Let's solve each part step by step:

### Problem 10: $ m\angle ABG $
The incenter $ G $ bisects $ \angle ABC $. From the diagram, $ \angle ABG $ is given as $ 20^\circ $? Wait, no—wait, the diagram shows $ \angle ABG $ is marked with $ 20^\circ $? Wait, maybe the angle at $ B $ is split into two equal parts. Wait, actually, looking at the diagram, $ \angle ABG $ is given as $ 20^\circ $? Wait, no—wait, the incenter bisects the angle, so if $ \angle ABC $ is split into two equal angles by $ BG $, then $ m\angle ABG = 20^\circ $? Wait, maybe the diagram shows $ \angle ABG = 20^\circ $. Wait, maybe I misread. Wait, the diagram has $ \angle ABG $ marked with $ 20^\circ $? Wait, no—wait, the problem says "use your notes on incenters". Wait, maybe the angle $ \angle ABC $ is $ 40^\circ $ (since $ BG $ bisects it, so $ \angle ABG = \angle GBC = 20^\circ $). So $ m\angle ABG = 20^\circ $. Wait, maybe the diagram shows $ \angle ABG = 20^\circ $. Let's assume that. So:

$ m\angle ABG = 20^\circ $

### Problem 11: $ m\angle BCA $
The incenter $ G $ bisects $ \angle BCA $. From the diagram, $ \angle GCA = 11^\circ $, so $ \angle BCA = 2 \times 11^\circ = 22^\circ $.

**Step 1:** Recognize that $ CG $ bisects $ \angle BCA $, so $ \angle BCA = 2 \times \angle GCA $.
**Step 2:** Substitute $ \angle GCA = 11^\circ $:
$ m\angle BCA = 2 \times 11^\circ = 22^\circ $

### Problem 12: $ m\angle BAC $
The sum of angles in a triangle is $ 180^\circ $. We know $ \angle ABC = 2 \times 20^\circ = 40^\circ $ (since $ BG $ bisects it) and $ \angle BCA = 22^\circ $ (from Problem 11). So:

**Step 1:** Recall $ \angle BAC + \angle ABC + \angle BCA = 180^\circ $.
**Step 2:** Substitute $ \angle ABC = 40^\circ $ and $ \angle BCA = 22^\circ $:
$ \angle BAC = 180^\circ - 40^\circ - 22^\circ = 118^\circ $? Wait, no—wait, maybe I made a mistake. Wait, if $ \angle ABG = 20^\circ $, then $ \angle ABC = 2 \times 20^\circ = 40^\circ $. Then $ \angle BAC = 180 - 40 - 22 = 118^\circ $? Wait, that seems large. Wait, maybe the angle $ \angle ABG $ is $ 20^\circ $, so $ \angle ABC = 40^\circ $, $ \angle BCA = 22^\circ $, so $ \angle BAC = 180 - 40 - 22 = 118^\circ $. But let's check again. Wait, maybe the diagram has $ \angle ABG = 20^\circ $, $ \angle GCA = 11^\circ $, so:

$ m\angle BAC = 180^\circ - (2 \times 20^\circ) - (2 \times 11^\circ) = 180 - 40 - 22 = 118^\circ $

### Problem 13: $ m\angle BAG $
The incenter $ G $ bisects $ \angle BAC $, so $ \angle BAG = \frac{1}{2} \angle BAC $. From Problem 12, $ \angle BAC = 118^\circ $, so:

**Step 1:** Recognize $ AG $ bisects $ \angle BAC $, so $ \angle BAG = \frac{1}{2} \angle BAC $.
**Step 2:** Substitute $ \angle BAC = 118^\circ $:
$ m\angle BAG = \frac{1}{2} \times 118^\circ = 59^\circ $

### Problem 14: $ DG $
The incenter is equidistant from all sides of the triangle. $ DG $ is the distance from $ G $ to $ AB $, and $ GE $ (or $ GF $) is the distance to $ AC $ or $ BC $. From the diagram, $ GE = 4 $, so $ DG = 4 $ (since incenter is equidistant from all sides).

### Problem 15: $ BE $
$ BE $ is a segment from $ B $ to $ E $ (on $ AC $). From the diagram, $ BD = 11 $ (since $ DG $ is perpendicular to $ AB $, and $ BD = 11 $), and since the incenter is equidistant, $ BE = BD = 11 $? Wait, no—wait, $ BD $ and $ BE $ are both tangents from $ B $ to the incircle, so they are equal. So $ BE = BD = 11 $.

### Problem 16: $ BG $
To find $ BG $, we can use the Pythagorean theorem in $ \triangle BDG $. $ BD = 11 $, $ DG = 4 $, so:

**Step 1:** In $ \triangle BDG $, $ BG^2 = BD^2 + DG^2 $.
**Step 2:** Substitute $ BD = 11 $, $ DG = 4 $:
$ BG^2 = 11^2 + 4^2 = 121 + 16 = 137 $
$ BG = \sqrt{137} \approx 11.70 $ (but maybe exact value is $ \sqrt{137} $)

### Problem 17: $ GC $
To find $ GC $, we can use the Pythagorean theorem in $ \triangle GEC $. $ GE = 4 $, $ EC = 20 $ (from the diagram), so:

**Step 1:** In $ \triangle GEC $, $ GC^2 = GE^2 + EC^2 $.
**Step 2:** Substitute $ GE = 4 $, $ EC = 20 $:
$ GC^2 = 4^2 + 20^2 = 16 + 400 = 416 $
$ GC = \sqrt{416} = \sqrt{16 \times 26} = 4\sqrt{26} \approx 20.396 $

### Final Answers:
10. $ \boldsymbol{20^\circ} $
11. $ \boldsymbol{22^\circ} $
12. $ \boldsymbol{118^\circ} $
13. $ \boldsymbol{59^\circ} $
14. $ \boldsymbol{4} $
15. $ \boldsymbol{11} $
16. $ \boldsymbol{\sqrt{137}} $ (or $ \approx 11.70 $)
17. $ \boldsymbol{4\sqrt{26}} $ (or $ \approx 20.40 $)

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 10: \( m\angle ABG \)

Problem 11: \( m\angle BCA \)

Problem 12: \( m\angle BAC \)

Problem 13: \( m\angle BAG \)

Problem 14: \( DG \)

Problem 15: \( BE \)

Answer:

Problem 10: \( m\angle ABG \)

Problem 11: \( m\angle BCA \)

Problem 12: \( m\angle BAC \)

Problem 13: \( m\angle BAG \)

Problem 14: \( DG \)

Problem 15: \( BE \)

Problem 16: \( BG \)

Problem 17: \( GC \)

Final Answers: