In a `DeltaPQR`. if `3 sinP +4 cosQ=6` and `4sinQ+3cosP=1`, then the angle `R` is equal to:

Options

A. `(5 pi)/6`
B. `pi/6`
C. `pi/4`
D. `(3pi)/4`

Explanation

Given in ` triangle PQR`
3 sin P + 4 cos Q =6
and 4 sin Q + 3 cos P =1
On squaring and adding the Eqs. (i) and (ii) we get
` 9 (sin^(2) P +cos^(2) P) +16 (sin^(2) Q + cos^(2) Q)`
` + 2 xx 3 xx 4 ( sin P cos Q +sin Q cos P ) =37`
` 24 [ sin (P +Q)] = 37 -25`
` sin (P +Q) = 1/2`
Since P and Q are angles of ` DeltaPQR`
` 0^(@) lt P, Q lt 180^(@)`
` P +Q = 30^(@) or 150^(@)`
` Rightarrow R =150^(@) or 30^(@)` [ respectively]
Hence, two cases arise here
Case I when R = ` 150^(@)`
` Rightarrow P +Q = 30^(@) Rightarrow 0 lt P , Q lt 30^(@)`
` Rightarrow sin P lt 1/2 , cos Q lt 1`
` Rightarrow 3 sin P + 4 sin Q lt 3/2 + 4`
`Rightarrow 3 sin P + 4 cos Q lt 11/2 lt 6`
` Rightarrow 3 sin P + 4 cos Q =6` , which is not possible
Case II when R `= 30^(@)`
Hence, R =` 30^(@)` is the only possibility