Question

ADC is a straight line. Find the unknowns following figures.

Answer 1

Method 1:

`\sf\\(i)\ \angle BCD=\angle CBD=28^o\ \ \ [\textsf{Base angles of isosceles triangle are equal.}]\\\\(ii)\ \angle ADB=\angle CBD+\angle BCD\ \ \ [\textsf{An exterior angle of a triangle is equal to the sum}\\\textsf{}\ \ \ \ \textsf{of the opposite interior angles.]}\\or,\ \angle ADB=28^o+28^o\\or,\ \angle ADB=56^o\\`

`\sf\\(iii)\ \angle BAD=\angle ABD=x\ \ \ [\textsf{AD=BD, base angles of isosceles triangle are equal.}]\\\\(iv)\ \angle BAD+\angle ABD+\angle ADB=180^o\\or,\ x+x+56^o=180^o\\or,\ 2x=124\\or,\ x=62^o`

Method 2:

`\sf\\(i)\ \angle BAD=\angle ABD=x\ \ \ [\textsf{Base angles of isosceles triangle are equal.}]\\\\(ii)\ \angle BDC=\angle BAD+\angle ABD\ \ \ [\textsf{An exterior angle of a triangle is equal to the sum}\\\textsf{}\ \textsf{}\ \textsf{}\ \ \ \textsf{of the opposite interior angles.]}\\or,\ \angle BDC=x+x\\or,\ \angle BDC=2x`

`\sf\\(iii)\ \angle BCD=\angle CBD=28^o\ \ \ \textsf{[Base angles of isosceles triangle are equal.]}\\\\(iv)\ \angle BDC+\angle CBD+\angle BCD=180^o\ \ \ [\textsf{Sum of angles of triangle is 180}^o.]\\or,\ 2x+28^o+28^o=180^o\\or,\ 2x=124^o\\or,\ x=62^o`

Method 3:

`\sf\\(i)\ \triangle ABC\textsf{ is right triangle.\ \ \ [AD=CD=BD, midpoint of hypotenuse being }\\\textsf{}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \textsf{equidistant to the vertices.]}\\or,\ \angle B=90^o\\\\(ii)\ \angle BCD=\angle CBD=28^o\ \ \ \textsf{[Base angles of isosceles triangle are equal.]}\\\\(iii)\ \angle ABD=x=\angle B-\angle CBD=90^o-28^o\\or,\ x=62^o`

Answer 2

x = 62°

Explanation:

Triangle BCD and ABD are both isosceles triangles so that means that Angle C is EQUAL to Angle B reason being (base angles of an isos. triangle.

So then...

28°+28°+ Angle D= 180°

56°+Angle D= 180°

Angle D = 180°- 56°

Angle D = 124°

Then...

The top angle of Triangle is equal to 56°

180° - 56° = 124°

124°÷ 2 = 62°

x = 62°

Angle A = 62°