In the given figure, \[CD||AB.\]Find y. |
A) \[79{}^\circ \]
B) \[72{}^\circ \]
C) \[74{}^\circ \]
D) \[74{}^\circ \]
Correct Answer: B
Solution :
In \[\Delta ABC,\]\[\angle ABC+\angle BCA+\angle CAB=180{}^\circ \] |
\[\Rightarrow \]\[4x+3x+3x=180{}^\circ \]\[\Rightarrow \]\[x=18{}^\circ \] |
\[AB||CD\]and \[BC\]is the transversal, then |
\[\angle ABC+\angle BCD=180{}^\circ \] |
[sum of the internal angles on the same side of the transversal] |
\[\Rightarrow \]\[\angle ABC+\angle BCA+\angle ACD=180{}^\circ \] |
\[\Rightarrow \] \[4x+3x+\angle ACD=180{}^\circ \] |
\[\Rightarrow \] \[\angle ACD=180{}^\circ -7x\] |
\[\Rightarrow \]\[\angle ACD=180{}^\circ -7\times 18{}^\circ =54{}^\circ \] |
Also, \[\angle ACE=\angle CAB+\angle ABC\]exterior angle is equal to the sum of interior opposite angles. |
\[\angle ACD+\angle DCE=3x+4x=7x\] |
\[\Rightarrow \] \[54{}^\circ +y=7\times 18{}^\circ =126{}^\circ \] |
\[\Rightarrow \] \[y=126{}^\circ -54{}^\circ =72{}^\circ \] |
You need to login to perform this action.
You will be redirected in
3 sec