Question

Ajutor!!
50 de puncte

Answer 1

Răspuns:

1.a

2cos30-tg60+√2cos45°=

2*√3/2-√3+√2*√2/2=

√3-√3+2/2=0+1=1

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b) sin90°+tg 45°/sin60°=

1+1/(√3/2)=1+2/√3=amplifici  cu  √3=

1+2√3/3

2-cos15=(√6+√2)/4

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Aplici  formula

cosx=cos²/2-sim²x/2

cos30=cos²15°-sin²15=√3/2

cos²15°-(1-cos²15)=√3/2

cos²15°-1+cos²15°=√3/2

2cos²15=√3/2+1

2cos²15=(√3+2)/2

cos²15=(√3+2)/4

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3)sin²30°+cos²60°=

(1/2)²+(√3/2)²=

1/4+3/4=1

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4)Aplici teorema  cosinusului

BC²=AB²+AC²-2AB*AC*cosA

2²=3²+4²-2*3*4cosA

4=9+16-24cos<A

4=25-24cos<A

4-25= -24cos<A

-21=-24 cos<A

cos<A= -21/24

cos <A= -7/8

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5.Aplici  teorema  sinudsului

BC=a  R=raza

a/sinA=2R

7√3/sin60=2R

7√3/(√3/2)=2R

14√3/√3=2R

2R=14

R=7

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6.sinx =5/13

Aplici formula

sin²x+cos²x=1

(5/13)²+cos²x=1

25/169+cos²x=1

cos²x=1-25/169

cos²x=144/169

cosx=√144/169=±12/13

Deoarece  x∈(π/2,π) se  ia  valoarea  negativa

cos x= -12/13

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