Question

What are expressions for MN and LN? Hint: Construct the altitude from M to LN.

MN =

(Type an exact answer, using radicals as needed.)

Answer 1

MN= x
`√(2)`

LN=
`(√(2x) )/(2)` +
`(√(6x) )/(2)`

Explanation:

Answer 2

Part 1)
`MN=x√(2)\ units`

Part 2)
`LN=x(√(2))/(2)(1+√(3))\ units`

Explanation:

The picture of the question in the attached figure

step 1

Find the measure of ang;e M

Remember that the sum of the interior angles in any triangle must be equal to 180 degrees

so

`M+L+N=180^o`

substitute the given values

`M+45^o+30^o=180^o`

`M=105^o`

step 2

Applying the law of sines find the length side MN

`(MN)/(sin(45^o))=(x)/(sin(30^o))`

Remember that

`sin(45^o)=(√(2))/(2)`

`sin(30^o)=(1)/(2)`

substitute

`(MN)/((√(2))/(2))=(x)/((1)/(2))`

`MN=x√(2)\ units`

step 3

Find the length side LN

Construct the altitude from M to LN.

In the right triangle of the left

`cos(45^o)=(d_1)/(x)` ---> by CAH (adjacent side divided by the hypotenuse)

Remember that

`cos(45^o)=(√(2))/(2)`

substitute

`(√(2))/(2)=(d_1)/(x)`

`d_1=x(√(2))/(2)\ units`

In the right triangle of the right

`tan(30^o)=(d_1)/(d_2)` ---> by TOA (opposite side divided by adjacent side)

Remember that

`tan(30^o)=(√(3))/(3)`

`d_1=x(√(2))/(2)\ units`

substitute

`(√(3))/(3)=(x(√(2))/(2))/(d_2)`

`d_2=(3x√(2))/(2√(3))\ units`

simplify

`d_2=(x√(6))/(2)\ units`

Find the length side LN

Remember that

`LN=d_1+d_2`

substitute the values

`LN=(x(√(2))/(2)+(x√(6))/(2))\ units`

simplify

`LN=x(√(2))/(2)(1+√(3))\ units`