In triangleabc,
3sina + 4cosb = 6
4sinb + 3cosa = 1
find the measure of anglec.

π/6

Step-by-step explanation:

Let's start to solve by finding the square roots of both functions.

the suqare root of 3sinA+4cosB=6 will be 9sin2A+24sinAcosB+ 16cos2B=36

the square root of 4sinB + 3cosA = 1 will be 16sin2B+24sinBcosA+9cos2A

Then let's sum both functions side by side and group them in appropriate variables:

sin2A+cos2B=1 (in order trigonometric rules)

sinAcosB+cosAsinB= sin(A+B) then:

9(sin2A+cos2B)+24(sinAcosB+cosAsinB)+16(sin2A+cos2B)=37

9+16+24sin(A+B)=37

sin(A+B)=1/2

sin(A+B)=sinπ/6

A+B=π/6 ==> π−π/6=5π/6

If we know A,B and C are angles of a triangle,

Thus, A+B+C=π

⟹π/6+C=π==>5π/6+C=π ==>π/6

what is the question? i am guessing you are looking for the area.

to find an area, you have to do length x width.

in this case, it would be 12 ft x 4 ft = 48 sq. ft

∡ eia and ∡ kij are congruent

step-by-step explanation: