Prove that; tan20+4sin20= square root of3

tan( 20 ) + 4 Sin( 20 ) =

( Sin( 20 ) / Cos( 20 ) ) + 4 Sin( 20 ) =

Sin( 20 ) + 4 Sin( 20 ).Cos( 20 ) / Cos( 20 ) =

Sin( 20 ) + 2 × 2 Sin(20).Cos(20)/ Cos(20) =

Sin( 20 ) + 2 × Sin( 40 ) / Cos( 20 ) =

Sin( 20 ) + 2Sin( 40 ) / Cos( 20 ) =

Sin( 20 ) + 2Cos( 50 ) / Cos ( 20 ) =

Sin( 20 ) + 2Cos( 20 + 30 ) / Cos( 20 ) =

2 × Cos( 30 + 20 ) =

2 × [ Cos(30).Cos(20) - Sin(30).Sin(20) ] =

2 × [ √3/2 × Cos(20) - 1/2 × Sin(20) ] =

√3 Cos(20) - Sin(20)

Sin( 20 ) + 2Cos ( 20 + 30 ) / Cos( 20 ) =

Sin( 20 ) + √3 Cos(20) - Sin(20) / Cos(20) =

Sin(20) - Sin(20) + √3 Cos(20) / Cos(20) =

0 + √3 Cos(20) / Cos(20) =

√3 Cos(20) / Cos(20) =

Cos(20) simplifies from the numerator and denominator of fraction

√3 × 1 / 1 =

√3

And we're done

no he doesnt because "-16' is negatiive

step-by-step explanation:

(2x + 16) + (x) = 180

step-by-step explanation:

given,

abcd is a quadrilateral inscribed in a circle

angle a is labeled x plus 16, (x + 16)

angle b is labeled x degrees, (x)

angle c is labeled 6x minus 4, (6x – 4)

angle d is labeled 2x plus 16, (2x + 16)

what equation can be used to solve for angle b?

we know that the sum of the opposite angles of a simple quadrilateral inscribed in a circle is 180 degrees.

angle b is opposite angle d

therefore, angle b + angle d = 180

(x) + (2x + 16) = 180 

same as (2x + 16) + (x) = 180