Find cos(20°)+2cos^2 (80°)
me pls.

i did not understand your question

Step-by-step explanation:

1

Step-by-step explanation:

2cos²(80°)=1+cos(2(80°))

From trigonometry identity that states: cos2θ=2cos²θ-1

And from that we have: 2cos²θ=1+cos2θ

which makes 2cos²(80°)=1+cos(2(80°))

                                      =1+cos(160°)

From the quadrant table, cosine is negative on the second quadrant i.e.

cos(180°-θ)=-cosθ

Therefore, cos(160°)=cos(180°-20°)=-cos(20°)

Going back to the original question,

 cos(20°)+2cos²(80°)=cos(20°)+1+cos(160°)

                                 =cos(20°)+1+(-cos(20°))

                                 =cos(20°)+1-cos(20°)

                                 =cos(20°)-cos(20°)+1

                                 =1.

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐

circumference divided by 3.141592654 then timesthe radius twice

step-by-step explanation:

circumference divided by 3.141592654 then timesthe radius twice