Question

BCD is a straight line. Find the unknowns in the following figures.

Answer 1

x = 63
y = 180 - 63 * 2 = 54

Answer 2

`\sf\\\ (i)\ \angle ACD=\angle ABC+\angle BAC\ \ \ [\textsf{An exterior angle of a triangle is equal to the sum of}\\\textsf{its two opposite non-adjacent interior angles.]}\\or,\ x=28^o+35^o\\\therefore\ x=63^o`

`\sf\\(ii)\ \angle ACD=\angle ADC=x=63^o\ \ \ \textsf{[Base angles of isosceles triangle are equal.]}\\\\(iii)\ \angle CAD+\angle ADC+\angle ACD=180^o\ \ \ \textsf{[Sum of angles of triangle is 180}^o.]\\or,\ y+x+x=180^o\\or,\ y+63^o+63^o=180^o\\or,\ y=54^o`

Alternative method:

`\sf\\(i)\ \angle ACD=\angle ADC=x\ \ \ \textsf{[Base angles of isosceles triangle are equal.]}\\\\(ii)\ \angle ABD+\angle BAD+\angle ADB=180^o\ \ \ [\textsf{Sum of angles of triangle is 180}^o.]\\or,\ 28^o+y+35^o+x=180^o\\or,\ x+y=117^o......(1)`

`\sf\\(iii)\ \angle ACD+\angle ADC+\angle CAD=180^o\ \ \ [\textsf{Sum of angles of triangle is 180}^o.]\\or,\ x+x+y=180^o\\or,\ 2x+y=180^o.....(2)`

`\sf\\\textsf{Subtracting equation (2) from (1),}\\+(2x+y)=180^o\\-(x+y)=-117^o\\or,\ x=63^o\\\\\textsf{Equation (1) becomes}\\63^o+y=117^o\\or,\ y=54^o`