Q. 465.0( 3 Votes )

In the given figure, find 3 tan θ – 2 sin α + 4 cos α.


Answer :

First of all, we find the value of RS

In right angled ∆RQS, we have


(RQ)2 + (QS)2 = (RS)2


(8)2 + (6)2 = (RS)2


64 + 36 = (RS)2


RS =100


RS =±10 [taking positive square root, since side cannot be negative]


RS =10




Now, we find the value of QP


In right angled ∆RQP


(RQ)2 + (QP)2 = (RP)2


(8)2 + (QP)2 = (17)2


64 + (QP)2 = 289


(QP)2 =289–64


(QP)2 =225


QP =225


QP =±15 [taking positive square root, since side cannot be negative]


QP =15


tan θ


Now, 3 tan θ – 2 sinα + 4cos α







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