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= 1/2 cl.

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 .

m∠CLN=3m∠ALM

Let m∠ALM = x so m∠CLN = 3x

According to the diagram,

m∠A = 90 - x and m∠c= 90 - 3x

As we know that sum of all the angles of a triangle is 180,

m∠A + m∠B + m∠C = 180

(90 - x) + 40 + (90 - 3x) = 180

220 - 4x = 180

-4x = -40

x = 10

So m∠A = 90 - x = 90 - 10 =80

m∠C = 90 - 3x = 90 - 30 = 60

In ΔCLN, ∠N= 90°, ∠C = 60° and ∠CLN = 30°

So by 30 -60-90 theorem

Side opposite to 90° = 1/2( side opposite to 30°)

CN = 1/2 (CL)

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!