In Triangle ABC angles A=44,B=81,c=55,radius of incircle=2.6 find sides AB,BC,AC & construct Triangle & circle

Answer 1

AB = r(cot(A/2) + cot(B/2))
BC = r(cot(B/2) + cot(C/2))
CA = r(cot(C/2) + cot(A/2))

Answer 2

Construct a triangle A'B'C' having angles A' = 44, B'= 81 and
C' = 55 degrees. Draw the bisectors of each angle which will be concurrent at point I, the incenter of triangle A'B'C'. Now draw a circle with radius 2.6 inches and center I. Draw perpendiculars from I to AC at D, to BC at E, and to AB at F. Draw a line || to A'B' and tangent to circle I. Draw similar lines || to B'C' and A'B'. Label their intersections as A,B,C. The triangle ABC is similar to triangle A'B'C' and is the required triangle.

The sides can now be computed by realizing that the length of each tangent from point A to circle I is 2.6cot 22. Similarly the other tangent lengths are 2.6cot 27.5 and 2.6cot 40.5.

The area of the triangle is 2.6s where s is the semiperimeter of the triangle.

Answer 3

The sum of the angles is a hundred and eighty°. 3 + 4 + 5 =12 a hundred and eighty ÷ 12 = 15 3 x 15 = 40 5, 4 x 15 = 60, 5 x 15 = seventy 5 The angles of the triangle are 40 5°, 60° and seventy 5°. permit s, m and l denote the smallest, center and longest facets respectively. by the sine rule: s/sin(40 5) = m/sin(60) = l/sin(seventy 5) s/(½?2) = m/(½?3) = l/((a million+?3)/(2?2)) So the ratio of the facets is: ½?2 : ½?3 : (a million + ?3)/(2?2) Multiply by 2?2. 2 : ?6 : a million + ?3

Answer 4

AB = r(cot(A/2) + cot(B/2))
BC = r(cot(B/2) + cot(C/2))
CA = r(cot(C/2) + cot(A/2))

Answer 5

And what if i don't want to????