Question

AB = BC Angle BDC = 37½° CBD =

Answer 1

I hope this helps you

Answer 2

Answer:

$\angle CBD=105^{\circ}$

Step-by-step explanation:

We have been given a diagram. We are asked to find the measure of angle CBD using given diagram.

We can see that in triangle ABD two angles measure 60 degree. Using angle sum property of triangle, we can find measure of triangle angle ADB as:

$\angle A+\angle B+\angle D=180^{\circ}$

$60^{\circ}+60^{\circ}+\angle D=180^{\circ}$

$120^{\circ}+\angle D=180^{\circ}$

$120^{\circ}-120^{\circ}+\angle D=180^{\circ}-120^{\circ}$

$\angle D=60^{\circ}$

Since all angles of triangle ABD are equal, so it is an equilateral triangle and its all sides will have same measure.

$AB=BD=AD$

We have been given segment AB is equal to segment BC. Now, we will get:

$AB=BD=AD=BC$

In triangle BCD sides two sides (BD and BC) are equal, so it is an isosceles triangle.

We know that angles corresponding to equal sides of an isosceles triangle have equal measure, so measure of angle BDC will be equal to angle BCD.

$\angle BDC=\angle BCD=37\frac{1}{2}^{\circ}$

Now, we will use angle property of triangle to find measure of angle CBD as:

$\angle CBD+\angle BCD+\angle BDC=180^{\circ}$

$\angle CBD+37\frac{1}{2}^{\circ}+37\frac{1}{2}^{\circ}=180^{\circ}$

$\angle CBD+\frac{75}{2}^{\circ}+\frac{75}{2}^{\circ}=180^{\circ}$

$\angle CBD+\frac{150}{2}^{\circ}=180^{\circ}$

$\angle CBD+\frac{150}{2}^{\circ}-\frac{150}{2}^{\circ}=180^{\circ}-\frac{150}{2}^{\circ}$

$\angle CBD=\frac{360}{2}^{\circ}-\frac{150}{2}^{\circ}$

$\angle CBD=\frac{210}{2}^{\circ}$

$\angle CBD=105^{\circ}$

Therefore, the measure of angle CBD is 105 degrees.