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/cosx -5/cosy +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
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