PLEASE HELP ME! :)

1)
Given: △KLM
LM=12, m∠K=60°,m∠M=45°
Find: Perimeter of △KLM.

2) Given: △MNO
m∠M=45°, m∠O=30°
MN=6
Find: NO, MO.

PLEASE HELP ME 1 Given KLM LM12 mK60mM45 Find Perimeter of KLM 2 Given MNO mM45 mO30 MN6 Find NO MO class=
PLEASE HELP ME 1 Given KLM LM12 mK60mM45 Find Perimeter of KLM 2 Given MNO mM45 mO30 MN6 Find NO MO class=

Respuesta :

NO = 6√2

MO = 3√2+3√6

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

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