In the triangle ABC, it is known that the angle A is 40 °, the angle B is 20 °, AB – BC = 4 cm.

In the triangle ABC, it is known that the angle A is 40 °, the angle B is 20 °, AB – BC = 4 cm.Find the bisector of the triangle drawn from the vertex C.

Given: ΔАВС, SK – bisector; ∠ CAB = 40 °, ∠ ABC = 20 °, AB -BC = 4.
Find: SK -?
Solution:
Consider ΔАВС: by the property of angles ∠АСВ = 180 ° – 40 ° – 20 ° = 120 °, because SC – bisector, then ∠ ВСК = ∠ АСК = 60 °.
Let BC = x; AB = x + 4.
By the sine theorem BC / sin ∠А = AB / sin ∠ C;
x / 0.64 = (x + 4) / 0.87;
0.87x = 0.64 x + 2.56;
0.23 x = 2.56;
x = 11.1; BC = 11.1 cm; AB = BC + 4 = 15.1 cm.
Consider ΔВСК: ∠ВКС = 180 ° – 20 ° – 60 ° = 100 °.
By the sine theorem:
BC / sin 100 = CK / sin 20;
11.1 / 0.98 = x / 0.34;
0.98 x = 11.1 * 0.34;
0.98x = 3.77;
x = 3.77 / 0.98 = 3.85.
SK = 3.85 cm.
Answer: 3.85 cm.



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