No one ever answers my questions i need ! 2 problems given: △klm lm=12, m∠k=60°, m∠m=45° find: perimeter of △klm. and given: △mno m∠m=45°, m∠o=30° mn=6 find: no, mo.

To solve this problem we will use sinus theorem.

It's general form is

a/sinα = b/sinβ = c/sinγ  

In this case it reads ML/sin∡K = KL/sin∡M = KM/sin∡L  =>

ML/sin∡K = KL/sin∡M  =>  12/sin60° = KL/sin45° => 12/(√3/2) = KL/(√2/2) =>

2*12/√3 = 2*KL/√2 => We will divide whole equation with number 2 and get

12/√3=KL/√2 => KL*√3=12√2 => KL=(12√2)/√3

When we rationalize denominator we get  KL=4√6≈ 4*2.45= 9.8

Considering that we know two angles we will calculate the third

∡K+∡M+∡L=180° => 60°+45°+∡L=180° => ∡L= 180-105=75°

Reason- The theorem of the sum of the inner angles of the triangle

ML/sin∡K=KM/sin∡L => 12/sin60°=KM/sin75°

We will calculate  sin75° = sin (30+45)= sin30 *cos45+cos30*sin45 =>

sin75°= 1/2 * √2/2 = √3/2 * √2/2= √2/4 *(√3+1)≈(1.41/4)*(1.73+1)=0.96

12/(√3/2)=KM/0.96 => 12*2/√3=KM/0.96 => 13.87=KM/0.96 =>

KM=13.87*0.96= 13.31

Perimeter of ΔKLM => P=KL+ML+KM= 12+9.8+13.31= 35.11

In the same way you can calculate perimeter of ΔMNO.

Good luck!!!