Answer:
A) [tex]m\angle DBC=65^o[/tex] and [tex]AC=20\ in[/tex]
B) [tex]m\angle DBC=65^o15'[/tex] and [tex]AC=20\ in[/tex]
Step-by-step explanation:
The picture of the question in the attached figure
Problem A
we know that
Triangle ABC has two equal sides
AB ≅
BC
so
Is an isosceles triangle
therefore
[tex]m\angle CAB=m\angle BCA[/tex]
Remember that
The altitude to the base of an isosceles triangle bisects the vertex angle. and bisects the base
In this problem segment BD represent the altitude to the base of an isosceles triangle ABC
so
[tex]m\angle DBC=\frac{1}{2}m\angle ABC[/tex]
[tex]AC=2AD[/tex]
Part 1) Find the measure of angle DBC
we have
[tex]m\angle ABC=130^o[/tex]
substitute
[tex]m\angle DBC=\frac{1}{2}130^o=65^o[/tex]
Part 2) Find the value of segment AC
we have
[tex]AD=10\ in[/tex]
substitute
[tex]AC=2(10)=20\ in[/tex]
Problem B
we know that
Triangle ABC has two equal sides
AB ≅
BC
so
Is an isosceles triangle
therefore
[tex]m\angle CAB=m\angle BCA[/tex]
Remember that
The altitude to the base of an isosceles triangle bisects the vertex angle. and bisects the base
In this problem segment BD represent the altitude to the base of an isosceles triangle ABC
so
[tex]m\angle DBC=\frac{1}{2}m\angle ABC[/tex]
[tex]AC=2AD[/tex]
Part 1) Find the measure of angle DBC
we have
[tex]m\angle ABC=130^o30'[/tex]
substitute
[tex]m\angle DBC=\frac{1}{2}130^o30'=65^o15'[/tex]
Part 2) Find the value of segment AC
we have
[tex]AD=10\ in[/tex]
substitute
[tex]AC=2(10)=20\ in[/tex]
