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Answer:
Step-by-step explanation:
Since m∠1 = m∠2, KF is angle bisector of ∠K
In the ΔKFM we have:
Using low of sines:
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Rewritten by : Barada
Consider the attached diagram. Segment KO is an altitude of the triangle, so KO ⊥ LM. ∠L has measure 180° -105° -30° = 45°, so ΔKOL is an isosceles right triangle.
If we let segment LO have measure 1, then KO also has measure 1 and KL has measure √(1²+1²) = √2 by the Pythagorean theorem.
ΔKMO is half of an equilateral triangle, so KM has measure 2, and MO has measure √(2²-1²) = √3 by the Pythagorean theorem.
Then the ratio of KM to LM is 2:(1+√3) and the ratio of KL to LM is √2:(1+√3). That is, ...
... KL = LM×(√2)/(1+√3) = (20√3)(√2)/(1 +√3)
... KL = (20√6)/(√3 +1) = (20√6)(√3 -1)/(3 -1) . . . . . with denominator rationalized
... KL = 30√2 -10√6 ≈ 17.9315
and
... KM = LM×2/(1+√3) = KL×√2
... KM = (30√2 -10√6)√2
... KM = 60 -20√3 ≈ 25.3590
== == == == == ==
The Law of Sines tells you ...
... KL/sin(M) = KM/sin(L) = LM/sin(K)
Then ...
... KL = sin(M)·LM/sin(K) = sin(30°)·20√3/sin(105°) ≈ 17.9315
... KM = sin(L)·LM/sin(K) = sin(45°)·20√3/sin(105°) ≈ 25.3590
