The two sides of an acute-angled triangle are 13 cm and 20 cm, respectively.

| May 25, 2021 | education

The two sides of an acute-angled triangle are 13 cm and 20 cm, respectively. The radius of the circle described about the triangle is 65/6 cm. Find the third side of the triangle.

Let ABC be a given triangle. AB = 13 cm, BC = 20 cm.

Let’s denote the middle of the side AB by the letter K, and the middle BC by the letter M. Draw perpendiculars through the points K and M, the intersection point O is the center of the circumscribed circle around the triangle ABC.

KM is the middle line of the triangle ABC (connects the midpoints of the sides AB and BC), which means that KM = 1/2 AC.

Triangle KBO is rectangular (KO is perpendicular to AB), OB is the radius of the circumscribed circle, OB = 65/6. KB = 1/2 AB = 13/2.

By the Pythagorean theorem: KO = √ ((65/6) ² – (13/2) ²) = 52/6 = 26/3.

The BMO triangle is rectangular (OM is prependicular to BC), ОВ = 65/6, ВМ = 1/2 ВС = 10.

By the Pythagorean theorem: ОМ = √ ((65/6) ² – 10²) = √ (635/36) = 25/6.

Let’s calculate the sines and cosines of the angles KBO and MBO:

cosКBО = (13/2) / (65/6) = 39/65.

cosМBО = 10 / (65/6) = 12/13.

sinКBО = (26/3) / (65/6) = 52/65.

sinМBО = (25/6) / (65/6) = 5/13.

The angle KBM is equal to KBO + MBO:

cos (α + β) = cosα * cosβ – sinα * sinβ.

cosKVM = (39/65) * (12/13) – (52/65) * (5/13) = 16/65.

By the cosine theorem:

КМ² = KB² + BM² – 2 * KB * BM * cosКBМ = (13/2) ² + 10² – 2 * 13/2 * 10 * 16/65 = 28665/260 = 441/4.

KM = √ (441/4) = 21/2 = 10.5.

Therefore, AC = 2 * KM = 2 * 10.5 = 21 (cm).



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