Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
$ABCD$ is a kite and $\angle A=\angle C$. If $\angle CAD=60^{\circ}$ and $\angle CBD=45^{\circ}$, find
$( a).\ \angle BCD$
$( b). \angle CDA$
"
Given: $ABCD$ is a kite and $\angle A=\angle C$ and $\angle CAD=60^{\circ}$ and $\angle CBD=45^{\circ}$.
To do: To find:
$( a).\ \angle BCD$
$( b).\ \angle CDA$
Solution:
Consider the point of intersection of the two diagonals as $O$.
In $\vartriangle AOD$ and $\vartriangle COD$,
$AD=CD$ [Adjacent sides of kite are equal]
$AO=CO$ [BD bisects AC]
$OD=OD [Common]
$\therefore \vartriangle AOD\cong\vartriangle COD$
Hence,
$\angle CAD=\angle ACD=60^o$
Using angle sum property in $\vartriangle ACD$,
$\angle CAD+\angle ACD+\angle CDA=180^o$
$\Rightarrow 60+60+\angle CDA=180$
$\Rightarrow \angle CDA=180-120$
$\Rightarrow \angle CDA=60^o$
In $\vartriangle AOB$ and $\vartriangle COB$,
$AO=CO$ [BD bisects AC]
$OB=OB$ [Common]
$AB=BC$ [Adjacent sides of kite are equal]
$\therefore \vartriangle AOB\cong\vartriangle COB$
Hence,
$\angle AOB=\angle COB$
$\angle BAO=\angle BCO$
Using linear pair property,
$\angle AOB+\angle COB=180^o$
$\Rightarrow 2\times\angle AOB=180$
$\Rightarrow \angle AOB=90^o=\angle COB$
Using angle sum property in $\vartriangle BOC$.
$\angle OBC+\angle COB+\angle BCO=180$
$\Rightarrow 45+90+\angle BCO=180$
$\Rightarrow \angle BCO=180-135$
$\Rightarrow \angle BCO=45^o$
Therefore,
$\angle BCD=\angle BCO+\angle DCO$
$\angle BCD=45+60$
$\angle BCD=105^o$
Updated on: 10-Oct-2022
212 Views
Advertisements