Question

Given that the perimeter of a triangle is 148. If the angles of the triangle are in the ratio 5:6:7, determine the longest side of the triangle.

Answer 1

the angles Since is 180° total sum of a triangle be sx, 60, 72 the angle of a triangle sxt 6x+ 7x=180° 37 18x = 180° x= 10° There fore the twee anyles often triangles (7x10") b 16oo (sx 10), (6x10) co 70° a, b, c be the sides of the triangle opposite sides of the angle 500, 60, too of the to langle respectively By low of sines singtos sien Goo sin sin com a singo singoo and atbtc= 188 ti a Sinsoo & B Sinbo sinto o (Sam) => K sin 500 K sinboo (= k sin too By k singoo tusinbooths in 700 = 148 ar K(sinsoo + sinboot sin 70°) 2198

ar KX 2.571762 468 = 148 57 548 0 8 302 Therefore a= K singoo 57. 54808302x Sinsoo 44.084 38921 a = 49.88 (Round two decimet place) ksin boo 57.54803028 singo = 49.83805609 => b= 49.84 (Roura two decimal places kin Foo 37.54 80 3028 Sinzoo 54.07745932 54.0 (round trodecinul plency Hine logest side of the triangle is 54.07795932 bit OR c Ñ 54 Un't OR c = 51.08 (roma two decimal please)