Question

directions: if g is the incenter of (\triangle abc), find each measure.

1. (mangle abg)
2. (mangle bca)
3. (mangle bac)
4. (mangle bag)
5. (dg)
6. (be)
7. (bg)
8. (gc)

Explanation:

Step1: Recall the property of the in - center

The in - center of a triangle is the intersection of the angle bisectors of the triangle.

Step2: Find \(m\angle ABG\)

Since \(G\) is the in - center and \(BG\) is an angle bisector. Given that \(m\angle GBC = 20^{\circ}\), then \(m\angle ABG=m\angle GBC = 20^{\circ}\)

Step3: Find \(m\angle BCA\)

Since \(G\) is the in - center and \(CG\) is an angle bisector. Given that the angle adjacent to \(\angle BCA\) (the angle formed by the extension of \(BC\) and \(AC\)) is \(11^{\circ}\), then \(m\angle BCA = 2\times11^{\circ}=22^{\circ}\)

Step4: Find \(m\angle BAC\)

We know that the sum of angles in a triangle is \(180^{\circ}\). Let \(m\angle ABG = 20^{\circ}\), \(m\angle BCA=22^{\circ}\). Then \(m\angle ABC=2\times20^{\circ} = 40^{\circ}\)
Using the angle - sum formula for \(\triangle ABC\): \(m\angle BAC=180^{\circ}-m\angle ABC - m\angle BCA\)
\(m\angle BAC=180-(40 + 22)=118^{\circ}\)

Step5: Find \(m\angle BAG\)

Since \(AG\) is an angle bisector of \(\angle BAC\) and \(m\angle BAC = 118^{\circ}\), then \(m\angle BAG=\frac{1}{2}m\angle BAC=\frac{1}{2}\times118^{\circ}=59^{\circ}\)

Step6: Find \(DG\)

Since \(G\) is the in - center, the lengths of the perpendiculars from \(G\) to the sides of the triangle are equal. Given \(GF = 4\) (perpendicular from \(G\) to \(AC\)), \(DG\) (perpendicular from \(G\) to \(AB\)) is also \(4\)

Step7: Find \(BE\)

Since \(G\) is the in - center and \(BE\) is a side - related length. Given \(BC\) has a segment of length \(11\) (from the congruent segments related to the in - center's perpendiculars and angle bisectors), \(BE = 11\)

Step8: Find \(BG\)

Using the Pythagorean theorem in \(\triangle BDG\). \(BD = 11\), \(DG = 4\)
\(BG=\sqrt{BD^{2}+DG^{2}}=\sqrt{11^{2}+4^{2}}=\sqrt{121 + 16}=\sqrt{137}\approx11.7\) (but if we consider the non - calculation based on the given figure's congruent segments and in - center properties, and assuming some integer - like value from the problem's context, if we consider the right - triangle with legs \(4\) and \(11\) - like in the figure's structure, but if we assume from the problem's numbering and given values, maybe there is a mis - understanding. However, if we consider the fact that \(BD = 11\), \(DG=4\), by Pythagoras \(BG=\sqrt{11^{2}+4^{2}}=\sqrt{121 + 16}=\sqrt{137}\). But if we consider the problem's hand - written answers (maybe a mis - take in the problem's figure interpretation, if we assume that \(BG\) is related to the side \(BE = 11\) (but that's not correct by the in - center and right - triangle properties). The correct formula is \(BG=\sqrt{BD^{2}+DG^{2}}\)

Step9: Find \(GC\)

Using the Pythagorean theorem in \(\triangle CEG\). \(CE = 20\), \(GE = 4\)
\(GC=\sqrt{CE^{2}+GE^{2}}=\sqrt{20^{2}+4^{2}}=\sqrt{400+16}=\sqrt{416}=4\sqrt{26}\approx20.4\) (but if we consider the problem's hand - written answers (maybe a mis - take in the problem's figure interpretation, if we assume that \(GC\) is related to the side \(CE = 20\) (but that's not correct by the in - center and right - triangle properties). The correct formula is \(GC=\sqrt{CE^{2}+GE^{2}}\)

Answer:

1. \(20^{\circ}\)
2. \(22^{\circ}\)
3. \(118^{\circ}\)
4. \(59^{\circ}\)
5. \(4\)
6. \(11\)
7. \(\sqrt{137}\)
8. \(4\sqrt{26}\)