Question

solve it :-

correctly ✌✌✌​

Answer 1

Answer:

ABC is right angle triangle at A

AC=?,AB=8,BC=17

BC^2=AC^2+AB^2

17^2=AC^2+8^2

289=AC^2+64

AC=15

cos (theta)=adj/hyp

cosB=8/17

TanC=opp/adj=8/15

Sin^2B+cos^2B=(15/17)^2 +(8/17)^2=225+64/289=1

sinB.cosC +cosB.sinC=(15/17)(15/17) +(8/17)(8/17)=225/289 +64/289=1

Answer 2

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answer referred to the attachment

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