In a right-angled triangle, one of the corners is 60 *, and the distance from the center
In a right-angled triangle, one of the corners is 60 *, and the distance from the center of the inscribed circle to the apex of this angle is 10 cm. Find the larger side of this triangle. help me please.
∠В = 60 °, ∠С = 90 °, BO = 10 cm. The circle touches the sides of the triangle AC, AB and BC at points M, K and T, respectively. The larger side of a right-angled triangle is its hypotenuse. B △ ABC hypotenuse – AB.
1. Segments of tangents to the circle drawn from one point are equal to each other and make equal angles with a segment drawn from this point to the center of the circle ⇒ BT = BK, ∠OBT = ∠OBK = x.
∠В consists of two angles ∠ОВТ and ∠ОВК, then:
∠ОВТ + ∠ОВК = ∠В;
x + x = 60 °;
2x = 60 °;
x = 60 ° / 2;
x = 30 °.
∠ОВТ = ∠ОВК = x = 30 °.
2. From the center of the circle, draw a segment OТ (radius) to the point of tangency with the side BC. OT is perpendicular to BC, since the radius drawn to the tangent point is perpendicular to the tangent. Consider △ OTB: ∠ OTB = 90 ° (since OT⊥BC), ∠ OTB = 30 °, OB = 10 cm – the hypotenuse (since it lies opposite the right angle). From the properties of a right-angled triangle, it is known that opposite an angle of 30 °, there is a leg, which is 2 times less than the hypotenuse, then: OT = OB / 2 = 10/2 = 5 (cm).
By the Pythagorean theorem:
ОВ² = OT² + ВТ²;
5² + BT² = 10²;
BT² = 100 – 25;
BT² = 75;
BT = √75;
BT = 5√3 cm.
3. From the center of the circle, draw a segment OM (radius) to the point of tangency with the AC side. Consider the СMOT quadrilateral: MС⊥OM, OT⊥OM, OT⊥TС, MС⊥СT – all sides of the СMOT quadrilateral are mutually perpendicular. Since OM = OT, then CMOT is a square ⇒ CM = OM = OT = СT = 5 cm.
Thus, side △ ABC BC consists of two segments:
BC = BT + CT;
BC = 5√3 + 5 cm.
4. Consider △ ABC. By the theorem on the sum of the angles of a triangle:
∠А + ∠В + ∠С = 180 °;
∠А + 60 ° + 90 ° = 180 °;
∠А = 180 ° – 150 °;
∠А = 30 °.
From the properties of a right-angled triangle, it is known that opposite an angle of 30 °, there is a leg, which is 2 times less than the hypotenuse, then:
BC = AB / 2 ⇒ AB = 2 * BC;
AB = 2 * (5√3 + 5) = 10√3 + 10 (cm).
Answer: AB = 10√3 + 10 cm.
