Lets see who solves this problem?

Question

In a Triangle PQR:

4sinP+3cosQ=6
4cosP+3sinQ=1

Find angle R

Answer 1

In triangle PQR, sum of angles is equals to 180 degrees, i.e, P + Q + R = 180.

The trigonometric equations are 4 sin P + 3 cos Q = 6 → (1)

                                                   4 cos P + 3 sin Q = 1 → (2)

Multiply equation (1) by equation (1) and equation (2) by equation (2).

Write the equations colmn form and then add them.

16 sin² P + 12 cos Q sin P + 9 cos² Q = 36

16 cos² P + 12 cos P sin Q + 9 sin² Q = 1

( + )____________________________________________

16(sin² P + cos² P) + 12(cos Q sin P + cos P sin Q) + 9(sin² Q + cos² Q) = 37

Pythagoras identity : sin² P + cos² P = 1.

Compound angle formula : sin P cos Q + cos P sin Q = sin(P + Q).

16(1) + 12sin (P + Q) + 9(1) = 37

25 + 12sin (P + Q) = 37

12sin (P + Q) = 37 - 25

12sin (P + Q) = 12

sin (P + Q) = 1

sin (180 - R) = 1         [Since, P + Q + R = 180]

sin R = 1

sin R = sin (π/2)

⇒ R = sin^{- 1} (sin (π/2))

R = π/2.

Therefore, angle R is π/2.