Question

use your notes on incenters to
directions: if g is the incenter of δa
diagram: triangle with vertices b, a, c; points d (on ba, right angle), e (on bc, right angle), f (on ac, right angle); g is the incenter. lengths: bd=11, ec=4, fc=20; angles: ∠gbc=20°, ∠gca=11°. questions: 10. m∠a... 12. m∠b... 14. dg 16. bg

Explanation:

To solve the problem related to the incenter \( G \) of \( \triangle ABC \), we analyze each part:

10. \( m\angle A \)

The incenter is the intersection of angle bisectors. For \( \angle C \), we know one part is \( 11^\circ \), so \( \angle C = 2 \times 11^\circ = 22^\circ \). For \( \angle B \), one part is \( 20^\circ \), so \( \angle B = 2 \times 20^\circ = 40^\circ \). Using the triangle angle sum (\( 180^\circ \)):
\[
m\angle A = 180^\circ - \angle B - \angle C = 180^\circ - 40^\circ - 22^\circ = 118^\circ
\]

12. \( m\angle BAC \) (Same as \( m\angle A \))

From above, \( m\angle BAC = 118^\circ \).

14. \( DG \)

The incenter is equidistant from all sides (inradius). \( GE = 4 \) (perpendicular to \( AC \)), so \( DG \) (perpendicular to \( AB \)) equals \( GE \):
\[
DG = 4
\]

16. \( BG \)

We use the right triangle \( \triangle BDG \). \( BD = 11 \), \( DG = 4 \). By the Pythagorean theorem:
\[
BG = \sqrt{BD^2 + DG^2} = \sqrt{11^2 + 4^2} = \sqrt{121 + 16} = \sqrt{137} \approx 11.7
\]

Final Answers:

1. \( \boldsymbol{118^\circ} \)
2. \( \boldsymbol{118^\circ} \)
3. \( \boldsymbol{4} \)
4. \( \boldsymbol{\sqrt{137} \approx 11.7} \) (or exact \( \sqrt{137} \))

Answer:

To solve the problem related to the incenter \( G \) of \( \triangle ABC \), we analyze each part:

10. \( m\angle A \)

The incenter is the intersection of angle bisectors. For \( \angle C \), we know one part is \( 11^\circ \), so \( \angle C = 2 \times 11^\circ = 22^\circ \). For \( \angle B \), one part is \( 20^\circ \), so \( \angle B = 2 \times 20^\circ = 40^\circ \). Using the triangle angle sum (\( 180^\circ \)):
\[
m\angle A = 180^\circ - \angle B - \angle C = 180^\circ - 40^\circ - 22^\circ = 118^\circ
\]

12. \( m\angle BAC \) (Same as \( m\angle A \))

From above, \( m\angle BAC = 118^\circ \).

14. \( DG \)

The incenter is equidistant from all sides (inradius). \( GE = 4 \) (perpendicular to \( AC \)), so \( DG \) (perpendicular to \( AB \)) equals \( GE \):
\[
DG = 4
\]

16. \( BG \)

We use the right triangle \( \triangle BDG \). \( BD = 11 \), \( DG = 4 \). By the Pythagorean theorem:
\[
BG = \sqrt{BD^2 + DG^2} = \sqrt{11^2 + 4^2} = \sqrt{121 + 16} = \sqrt{137} \approx 11.7
\]

Final Answers:

1. \( \boldsymbol{118^\circ} \)
2. \( \boldsymbol{118^\circ} \)
3. \( \boldsymbol{4} \)
4. \( \boldsymbol{\sqrt{137} \approx 11.7} \) (or exact \( \sqrt{137} \))