Answer:
1). Perimeter = 35.18 units
2). NO = 4.24 units, MO = 11.59 units
Step-by-step explanation:
1). Given : In ΔKLM,
LM = 12, m∠K = 60°, m∠M = 45°
Then we have to find the perimeter of the given triangle.
By applying sine rule in the triangle,
[tex]\frac{sinK}{LM}=\frac{sinL}{KM}=\frac{sinM}{KL}[/tex]
Since ∠K + ∠L + ∠M = 180°
60 + ∠L + 45 = 180
∠L + 105 = 180
∠L = 180 - 105
∠L = 75°
[tex]\frac{sin60}{12}=\frac{sin75}{KM}[/tex]
[tex]\frac{0.866}{12}=\frac{0.966}{KM}[/tex]
KM = [tex]\frac{0.966\times 12}{0.866}[/tex]
KM = 13.38 unit
Similarly, [tex]\frac{sinK}{LM}=\frac{sinM}{KL}[/tex]
[tex]\frac{sin60}{12}=\frac{sin45}{KL}[/tex]
KL = [tex]\frac{0.707\times 12}{0.866}[/tex]
KL = 9.80 unit
Therefore, Perimeter of the triangle = LM + KL+ KM
= 12 + 9.80 + 13.38
= 35.18 units
2). Given in the ΔMNO,
m∠M = 45°, m∠O = 30° and MN = 6 units
To Find : Measure of NO and MO
m∠M + m∠N + m∠O = 180
45 + m∠N + 30 = 180
m∠N + 75 = 180
m∠N = 180 - 75
m∠N = 105°
Now we apply the sine rule in the triangle.
[tex]\frac{sinM}{NO}=\frac{sinN}{MO}=\frac{sinO}{MN}[/tex]
[tex]\frac{sinN}{MO}=\frac{sinO}{MN}[/tex]
[tex]\frac{sin105}{MO}=\frac{sin30}{6}[/tex]
[tex]\frac{0.966}{MO}=\frac{0.5}{6}[/tex]
MO = [tex]\frac{6\times 0.966}{0.5}[/tex]
MO = 11.59 units
[tex]\frac{sinM}{NO}=\frac{sinO}{MN}[/tex]
[tex]\frac{sin45}{NO}=\frac{sin30}{6}[/tex]
NO = [tex]\frac{0.5\times 6}{0.707}[/tex]
NO = 4.24 units
