Question

In the given figure (not drawn to scale) ABCD is a rhombus and ALMO is a square, AC = BC Find
MBC
D​

Answer 1

Answer:

In the square ALMC, ∠ACM=90  

o

 

Also, AC=BC (Given)  

and BC=AB (Sides of Rhombus)

So, △ABC is an equilateral triangle  

Now, ∠ACB=60  

o

 

So from figure , ∠MCB=90  

o

−60  

o

=30  

o

 

Now in triangle MBC, BC=CM, so ∠MBC=∠BMC=x(let)

Sum of all angle of a triangle =180  

o

 

∠MCB+∠MBC+∠BMC=180  

o

 

⇒30  

o

+x+x=180  

o

 

⇒2x=180  

o

−30  

o

 

⇒x=75  

0

Step-by-step explanation: