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Question

In the following figure :
AD BC, AC = 26, CD = 10, BC = 42, DAC=x and B=y.
Find the value of :6cosx5cosy+8tany is m14, m is
196234_1d47cf0b207f4b13b621e00ac7215bbb.png

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Solution

Given: ABC, ADBC, AC=26, CD=10, BC=42
Now, In ACD,
AC2=CD2+AD2
262=102+AD2
AD2=576
AD=24
Now, In ABD,
BD=BCCD
BD=4210
BD=32
AB2=BD2+AD2
AB2=322+242
AB2=1600
AB=40
Now, 6cosx5cosy+8tany
= 6secx5secy+8tany
= 6(HB)5(HB)+8(PB)
= 6(ACAD)5(ABBD)+8(ADBD)
= 6(2624)5(4032)+8(2432)
= 132254+6
= 2625+244
= 254
= 614

m=6

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