Question

Given:

AB
≅
BC
and AD = 10 in
m∠BDC = 90°
m∠ABC = 130°
Find: m∠ DBC, AC

Answer 1

A)
`m\angle DBC=65^o` and
`AC=20\ in`

B)
`m\angle DBC=65^o15'` and
`AC=20\ in`

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

`m\angle CAB=m\angle BCA`

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

`m\angle DBC=(1)/(2)m\angle ABC`

`AC=2AD`

Part 1) Find the measure of angle DBC

we have

`m\angle ABC=130^o`

substitute

`m\angle DBC=(1)/(2)130^o=65^o`

Part 2) Find the value of segment AC

we have

`AD=10\ in`

substitute

`AC=2(10)=20\ in`

Problem B

we know that

Triangle ABC has two equal sides

AB ≅

BC

so

Is an isosceles triangle

therefore

`m\angle CAB=m\angle BCA`

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

`m\angle DBC=(1)/(2)m\angle ABC`

`AC=2AD`

Part 1) Find the measure of angle DBC

we have

`m\angle ABC=130^o30'`

substitute

`m\angle DBC=(1)/(2)130^o30'=65^o15'`

Part 2) Find the value of segment AC

we have

`AD=10\ in`

substitute

`AC=2(10)=20\ in`