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

Question

The question is-

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

Answer 1

....ok let's go...

Let's square both sides of both equations and combine like terms.

Squaring the first equation gives....................9Sin²(P) + 16Cos²(Q)+ 24Sin(P)Cos(Q)=36

Squaring the second equation gives...............16Sin²(Q) + 9Cos²(P) + 24Sin(Q)Cos(P)=1

Now, let's add the 2 equations...we get

9Sin²(P) + 16Cos²(Q)+ 24Sin(P)Cos(Q) + 16Sin²(Q) + 9Cos²(P) + 24Sin(Q)Cos(P)=37

Grouping the Sin²(P) and the Cos²(P) and Sin²(Q) and the Cos²(Q), we get

9Sin²(P) + 9Cos²(P) + 16Cos²(Q)+ 16Sin²(Q) + 24Sin(P)Cos(Q) + 24Sin(Q)Cos(P)=37

Now let's factor by grouping

9(Sin²(P) + Cos²(P) )+ 16(Cos²(Q)+ Sin²(Q) )+ 24(Sin(P)Cos(Q) + Sin(Q)Cos(P))=37

...and recall that Sin²(x) + Cos²(x) =1 and Sin(A+B)=Sin(A)Cos(B)+Sin(B)Cos(A)

9 + 16 + 24(Sin(P+Q))= 37

25 +24Sin(P+Q)=37

24Sin(P+Q)=12

Sin(P+Q) =1/2

No Recall Sin(30) =1/2 and the sum of angles in a triangle equals 180⁰

so if Sin(P+Q) =1/2

and Sin(30) =1/2

then P + Q = 30

Hence angle R =150⁰