Rectangle ABCD has AB = 6 and BC = 3. Point M is chosen on side AB so

First time posting - please excuse the format.

Let ∠AMC = ∠CMD = x,

From the triangle AMC, we know ∠ACM + ∠AMC + ∠MAC = 180, i.e. ∠ACM + x + 90 = 180 => ∠ACM = 90 - x

Since ABCD is a rectangle, we know that ∠ACM + ∠MCD is 90, which means that 90 - x + ∠MCD = 90 based on above.
This would mean ∠ MCD = 90- (90-x) = x

Now observe the triangle MCD, we already know ∠CMD is x from the question, we derived that ∠MCD is also x, which means triangle MCD is an isosceles triangle.
We now know CD = MD = 6. Also observe MD is the hypotenuse of triangle MBD.

Now consider the right-angled triangle MBD: MD = 6, BD = 3. This triangle follows the 1:\sqrt{3}:2 ratio (30:60:90). This would mean ∠BMD is 30.

From there we set up the formula ∠AMC + ∠CMD + ∠BMD = 180 => x+x+30= 180. => x = 75. Answer E.

Hope this helps.

Bogei

Attaching some images, self-reference solution that anyone can feel free to use.

The first image is the original question in illustrated form.



Based on this first image, we can come up with the second image which lets us fill in a lot of useful angle details. First, we get \(∠CMB = 180 - 2x\) because angles in a line must add up to 180. Then because angles in a triangle add up to 180, we can calculate that \(∠MCB = 180 - 90 - (180 - 2x) = 180 - 90 - 180 + 2x = 2x - 90\). Then because we know that all corners of a rectangle are 90 degrees, we can calculate that \(∠MCD = 90 - (2x - 90) = 90 - 2x + 90 = 180 - 2x\). Once again using the property of 180 degrees in a triangle, we can finally calculate that \(∠MDC = 180 - x - (180 - 2x) = 180 - x - 180 + 2x = x\).



This gives us useful information about the sides that we can see in the third image. Now that we know that triangle DMC has 2 of the same angles, it must be an isosceles triangle, and the sides opposite the equal angles are also equal in length. Since CD is 6 because it is a side of the rectangle, MC is also 6. Then, we can calculate that \(MB = \sqrt{6^2 - 3^3} = 3\sqrt{3}\) using the Pythagorean Theorem.



This triangle represents the properties of a 30-60-90 triangle, where the hypotenuse is \(2s\) (aka \(s = 3\)), the longer non-hypotenuse side across from the 60 degree angle is \(s\sqrt{3}\) (aka \(3\sqrt{3}\))and the shorter side opposite the 30 degree angle is s (aka \(3\)). Therefore, we know that ∠MCB is 60 degrees and ∠CMB is 30 degrees.

Finally, we can calculate for x. (you can use either ∠MCB or ∠CMB and it will produce the same result).

If using ∠CMB:
\(180 - 2x = 30\)
-> \(2x = 150\)
-> \(x = 75\)

If using ∠MCB:
\(2x - 90 = 60\)
-> \(2x = 150\)
-> \(x = 75\)

Answer choice E is correct. Hope this helps like it helped me!

Can someone tell how to solve this?

I got 60 which is I guess is the trap answer

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Sory guys , it is not giving me option to upload the picture which i drew



you can draw your self and assume

angle MCB = $

and follow the above mentioned approach

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Gmat doesnt expect us to use trigonometry....there must be another way to solve this

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Perfect analysis!



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Why angle AMC = angle CMD ??? They just give us that
angle AMD= angle CMD
can u explain

This does not require any Calculations,
If you were to just draw the figure , you will notice angle ∠BMC be Smaller than 90 Degrees.

Two possibilities are 30,60 ( 180 - 75*2 , 180 - 60*2 )

If you were to assume ∠BMC = 60
THEN BM = AM = sqrt{3}

However, AM = 6 Hence ∠BMC = 60 is rules out.

Hence ∠BMC = 30
angle AMD + angle CMD = 180 - 30
angle AMD = angle CMD = 150/2 = 75

This is my first time posting a reply. Please let me know if this is correct.

We know that angle AMD = angle DMC. Let's take them as = x.
angle AMD = angle MDC = x, as they are vertically opposite angles between parallel lines.
Therefore, DC = MC = 6.

In triangle BMC, by pythagoras theorem,
6^2 = 3^2 + BM^2
Thus, BM = 3*(3^(1/2))
By the property of right triangles, sides are in proportion of a 30:60:90 triangle. Thus angle BMC = 30.

So x + x + 30 = 180
Thus x = 75 i.e. E

­Here is how I did it.