If in the given figure AD is perpendicular to BC, AC = 26 units, CD = 10 units, BC = 42 units, triangle DAC whose angle is x and angle B = y, then, find the value of   6/cos⁡x -5/cos⁡y +8 tany =?           

In ∆ ADC –  AD =   √(AC)²-DC² )     AD =  √(26²-10² )  AD =   √(676-100)  AD = √576  =  24   AD   =   24   Now in ∆ABD  AB  = √(AD)²+(BD)² )   AB   =   √(24²+32²   AB   =    √(576+1024)    =  √1600   AB = 40 NOW-                                                       ∴The side opposite to the angle is the hypotenuse,  so the base for angle x will be AD. cosx=24/26  cosy=32/40 tany=24/32 6/cosx -5/cosy +8 tany =? =6/(24/26)-5/(32/40)+8×24/32 =26/4 - 25/4 + 24/4           =(50-25)/4   =25/4 6/cosx -5/cosy +8 tany =25/4