In a triangle
PQR, if 3sin P + 4cos Q = 6 and 4sinQ +3cos P =1, then the angle R is equal to
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Answer:
angle r is equal to 90 degree
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Step-by-step explanation:
9sin^2P+16cos^2Q+24sinPcosQ=36
9cos^2P+16sin^2Q+24sinQcosP=1
adding
9+16+24(sinPcosQ+sinQcosP)=37
24sin(P+Q)=37-25=12
sin(P+Q)=1/2
P+Q=30 or P+Q=150
R=180-30=150
or
R=180-150=30
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