LAW OF SINES
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[tex] \large \bold{\blue{Problem:}} [/tex] A triangular parcel of land with points M, K, L was to be fenced. No data for the lengths of sides MK and KL are available as shown in the figure.
- a) Find the distance between point K to point L.
- b) What is the distance from point M to K?
- c) How much fencing material is to be used in the lot?
[tex] \: [/tex]
[tex] \large \bold{\blue{Solution \ (a):}} [/tex] The given triangle is AAS (two angles and a side opposite to one angle). Find KL using the law of sines.
- [tex] \sf \frac{sin \ M}{KL} = \frac{sin \ K}{LM} \\ [/tex]
- [tex] \sf \frac{sin \ 52°}{KL} = \frac{sin \ 83°}{150m} \\ [/tex]
- [tex] \sf KL(sin \ 83°) = 118.2m [/tex]
- [tex] \sf \frac{KL(\cancel{sin \ 83°})}{\cancel{sin \ 83°}} = \frac{118.2m}{sin \ 83°} \\ [/tex]
- [tex] \sf KL = 119.09m [/tex]
[tex] \large \therefore \boxed{\tt \purple{KL = 119.09 \ meters}} [/tex]
[tex] \: [/tex]
[tex] \large \bold{\blue{Solution \ (b):}} [/tex] The given information is not enough. The opposite angle of side MK, Point L has no given measure of angle. To find it, remember that the sum of interior angles of a triangle is 180°.
- [tex] \sf \angle M + \angle K + \angle L = 180° [/tex]
- [tex] \sf 52° + 83° + \angle L = 180° [/tex]
- [tex] \sf 135° + \angle L = 180° [/tex]
- [tex] \sf \angle L = 180° - 135° [/tex]
- [tex] \sf \angle L = 45° [/tex]
Now that we have found the measure angle of point L, solve for MK using the law of sines.
- [tex] \sf \frac{sin \ L}{MK} = \frac{sin \ K}{LM} \\[/tex]
- [tex] \sf \frac{sin \ 45°}{MK} = \frac{sin \ 83°}{150m} \\ [/tex]
- [tex] \sf MK(sin \ 83°) = 106.07m [/tex]
- [tex] \sf \frac{MK(\cancel{sin \ 83°})}{\cancel{sin \ 83°}} = \frac{106.07m}{sin \ 83°}\\ [/tex]
[tex] \sf MK = 106.87m [/tex]
[tex] \large \therefore \boxed{\tt \purple{MK = 106.87 \ meters}} [/tex]
[tex] \: [/tex]
[tex] \large \bold{\blue{Solution \ (c):}} [/tex] To find how much fencing material is to be used, identify the perimeter of the triangle. Add all measure of the sides.
- [tex] \sf P = MK + KL + LM [/tex]
- [tex] \sf P = 106.87m + 119.09m + 150m [/tex]
- [tex] \sf P = 375.96m [/tex]
[tex] \large \therefore \boxed{\tt \purple{375.96 \ meters}} [/tex]
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