Question

Hello all!
Could someone please provide a step by step explanation for this sum?
the answer is supposed to be 75°

Answer 1

Given :-

• ∠LCB=15°

• ABCD is a rhombus

• ACML is a square

• AC=BC

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To find :-

• ∠MBC

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Solution :-

⇒ AC=BC [given]                       ...(1.)

⇒ AC=CM [Side of square]       ...(2.)

From equation (1) and (2),

⇒ BC=CM

Since BC=CM, ∆BCM will be an isosceles triangle.

⇒ ∠CMB = ∠CBM (Angle opposite to equal sides)  ...(3.)

Now, we can see from the given figure that CL is the diagonal of square. Diagonal of square divide both angles equally.

i.e. ∠ACL=∠LCM      ...(4.)

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Now we know that each angle of square is of 90°.

So,

⇒ ∠ACL+∠LCM=90°

⇒ ∠LCM+∠LCM=90°  [from equation (4.)]

⇒ 2∠LCM=90°

⇒ ∠LCM=90°/2

⇒∠LCM=45°

Now it is given that ∠LCB=15°  ...(5.)

⇒ ∠LCM=45°

⇒ ∠LCB+∠BCM=45°

⇒ 15°+∠BCM=45°    [From equation (5.)]

⇒ ∠BCM=45°-15°

⇒ ∠BCM=30°        ...(6.)

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Now applying angle sum property in ∆CBM

⇒ ∠MBC+∠CMB+∠BCM=180°

⇒ 30°+∠MBC+∠MBC=180°  [From equation (3.) and (6.)]

⇒ 2∠MBC+30°=180°

⇒ 2∠MBC=180°-30°

⇒ 2∠MBC=150°

⇒ ∠MBC=150°/2

⇒ ∠MBC=75°

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Answer 2

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In the square ALMC, ∠ACM=90°

Also, AC=BC (Given) 

and BC=AB (Sides of Rhombus)

So, △ABC is an equilateral triangle 

Now, ∠ACB=60°

So from figure , ∠MCB=90°−60°=30°

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

Sum of all angle of a triangle =180°

∠MCB+∠MBC+∠BMC=180°

⇒30°+x+x=180°

⇒2x=180°−30°

⇒x=750°

• I Hope it's Helpful my Friend.