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

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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