Given a triangle ABC, in it AH and BK heights that intersect at point O. Angle AOB is 120, OK is 3 and OB is 5.

| May 30, 2021 | education

Given a triangle ABC, in it AH and BK heights that intersect at point O. Angle AOB is 120, OK is 3 and OB is 5. Find AB and AC. It is very necessary, the grade for a quarter depends on this work.

Given a triangle ABC, in it AH and ВK are heights that intersect at point O, ∠AOB = 120 °, OK = 3 and OB = 5. Then ∠BOН = ∠AOK = 180 ° – ∠AOB = 180 ° – 120 ° = 60 ° and ∠OBН = ∠OAK = 90 ° – 60 ° = 30 °. In a right-angled triangle opposite an angle of 30 ° lies a leg equal to half of the hypotenuse, we obtain OH = 5: 2 = 2.5 (in Δ OHB, ∠ OHB = 90 °) and AO = 3 ∙ 2 = 6 (in Δ OKA, ∠ OKA = 90 °), and the height of the ВK = 3 + 5 = 8. In Δ OKA according to the Pythagorean theorem AO ^ 2 = AK ^ 2 + OK ^ 2; AK ^ 2 = 6 ^ 2 – 3 ^ 2 = 27 or AK = (27) ^ (1/2); then in Δ ВКА (∠BKA = 90 °) by the Pythagorean theorem AB ^ 2 = 27 + 8 ^ 2; AB ^ 2 = 91 or AB = 91 ^ (1/2) ≈ 9.5. In Δ BKС (∠BKС = 90 °), the angle BCК = 90 ° – 30 ° = 60 ° and tg60 ° = BK: KС, hence KС = 8: 3 ^ (1/2). AC = AK + KС = (27) ^ (1/2) + (8: 3 ^ (1/2)) = (17/3) ∙ 3 ^ (1/2) ≈ 9.8.
Answer: sides AB ≈ 9.5 and AC ≈ 9.8.



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