Please help me. Check if I completed the question correclty, with correct steps and answers. Image 1 is my workImage 2 is the question A 8 kg decoration hangs from the ceiling. It issuspended by 2 wires. The first wire forms a40° angle with the ceiling. The second wire forms a50° angle with the ceiling.a. Find the tension in the 2 wires (includingmagnitude and direction)b. Assuming the wire that forms the 40° angle withthe ceiling has a length of 0.3 metres, how farbelow the ceiling is the decoration suspended? #4wire # 2.0₁ = 40°0₂ = 50°10₂01a) Ti Cos 0₁ - T₂ cos A₂ = GF27 T₁ Cos 01 = T₂ Cos 0₂> T2Ti CostiCOS 02→TiTi =and Ti Sin 0₁ + T₂ Sinf₂ = • mgputting T₂[8 kg]T₁SindiTi sini +8₁(wire #1 - D Decoration Diagram8x9.8Sin 400+we are goingto use equation latercos erCos 0₂mgSin 01 + coS01c0382T₁ = 42, 735 N+ cos 40°Cos 500Ti CostiCos 0₂- mgCos 4000=and by 12 42.735 * cos 50°T₂ = 46.297 NX• mgFree Body Diagram2010Ti202T20₂ N101I cos 0₂ [Bjmy>l=0.3mT₁ = Sin Q₁ +T2 sin 02II COSE IApply Newtons second lawTI Cos 01 - T₂ cos 6₂ = mauTi sin 01 + T₂ Sing-mgAB6) From A ABC sing, IAB= lsing₁= 0.3x sin 40°AB= 0.193 m.= mayThe tension for wire 1 isT₁ = 42.735 N and wire #2its tension is 7₂ = 46.297 NXb) The distance for how farthe ceiling is where thedecoration is suspendedis AB = 0.193 m

Answered: Image /qna-images/answer/b7f14bb4-97aa-4c3d-a2f2-f1f048534db5.jpg

#4 10₁ = 40° 0₂ = 50° and wire #2 - 11 TIL Ti Sindi putting 1₂ O₂ Ti Cos T₂ cos 0₂ => T₁ Cos 0₁ = T₂ Cos 0₂ > T2₂ = Ti Costi COS 0₂ 0- and by Sin e T₁- = 8kg/m 1 m = Ti sin + + 0₁ (wire #1 - Decoration Diagram. + T₂ Sin02 = 6 T1= 42.735 N we are going to use equation latel cos i Cos 0₂ 8x9.8 T₁ = sin 40° + cos 40° Cos 500 mg Sin 01 + cos01 Cos 02 mg Ti Costi Cos 0₂ = mg 01₂ 42.735 * cos 50° Cos 400 X T2 = 46.297 N mg Free Body Diagram T 0₂ 72 F U 1 12 cos D₂ A 1 Ti 1 0₂ Nor B my 01 /C Apply Newtons second law TI Cos DI T₂ cos 62 = Mau Ti sin 1 + 1₂ sing-mg = may 2 AB b) From A ABC sing₁= I AB= lsing l=0.3m T₁=SnQ₁ + T2 sin 02 Ji Costi 0.3x sin 40° AB= 0.193 m a): The tension for wire I is T₁ = 42.735 N and wire #2 its tension is T₂ = 46.297 N b). The distance for how fa the ceiling is, where the decoration is suspended is AB = 0.193 m

#4 10₁ = 40° 0₂ = 50° and wire #2 - 11 TIL Ti Sindi putting 1₂ O₂ Ti Cos T₂ cos 0₂ => T₁ Cos 0₁ = T₂ Cos 0₂ > T2₂ = Ti Costi COS 0₂ 0- and by Sin e T₁- = 8kg/m 1 m = Ti sin + + 0₁ (wire #1 - Decoration Diagram. + T₂ Sin02 = 6 T1= 42.735 N we are going to use equation latel cos i Cos 0₂ 8x9.8 T₁ = sin 40° + cos 40° Cos 500 mg Sin 01 + cos01 Cos 02 mg Ti Costi Cos 0₂ = mg 01₂ 42.735 * cos 50° Cos 400 X T2 = 46.297 N mg Free Body Diagram T 0₂ 72 F U 1 12 cos D₂ A 1 Ti 1 0₂ Nor B my 01 /C Apply Newtons second law TI Cos DI T₂ cos 62 = Mau Ti sin 1 + 1₂ sing-mg = may 2 AB b) From A ABC sing₁= I AB= lsing l=0.3m T₁=SnQ₁ + T2 sin 02 Ji Costi 0.3x sin 40° AB= 0.193 m a): The tension for wire I is T₁ = 42.735 N and wire #2 its tension is T₂ = 46.297 N b). The distance for how fa the ceiling is, where the decoration is suspended is AB = 0.193 m