In △abc, m∠abc=40°, bl (l∈ac ) is the angle bisector of ∠ b. point m∈ab so that lm ⊥ab and n∈bc so that ln ⊥bc .find the measures of the angles of △abc if m∠cln=3m∠alm and prove that cn= 12 cl.

check the picture below.

M is perpendicular to AB and ∈ AB, and stemming from point L.

N is perpendicular to BC and ∈ BC, and stemming from point L as well.

BL is the bisector, that means the angle at verte B, gets cut into two equal halves.

now, we know what ∠CLN =3∠ALM, namely that ∠CLN is 3 times greater than ∠ALM.

so, once the bisector kicks in, you get two angles of 20° each, the angles at M are 90° each and the angles at N are 90° each as well, that pretty much narrows down what the missing angle is in triangles MBL and NBL, so is 70°.

now, the little sliver angles at CLN and ALM are on a 3:1 ratio, so, the flat-line of AC affords us 180°, subtract the 140°, so CLN and ALM will have to share only the remaining 40°, and they have to do so on a 3:1 ratio, that leaves us with, notice the blue angles.

now, from the triangles ALM and CLN, you can pretty much tell what the missing angle is, and those are the values for angles A and C.

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Solve :
 3 x 5 + 6 x 2 + 1 

Simplify promblem:

P.E.M.D.A.S's Rule

( 3 x 5 ) + ( 6 x 2 ) + 1
     V              V         V
    15      +    12   +   1
   {       27     +    1       }
                    V
                   28