The midpoint of $(\\var{xa1},\\var{ya1})$ and $(\\var{xa2},\\var{ya2})$ is $\\large($[[0]], [[1]]$\\large)$.

The midpoint of two points is simply the point whose $x$ coordinate is the average of the other $x$ coordinates, and whose $y$ coordinate is the average of the other $y$ coordinates. That is, the midpoint of $(\\var{xa1},\\var{ya1})$ and $(\\var{xa2},\\var{ya2})$ is the point $\\left(\\simplify[basic]{({xa1}+{xa2})/2},\\simplify[basic]{({ya1}+{ya2})/2}\\right)=\\left(\\simplify{({xa1}+{xa2})/2},\\simplify{({ya1}+{ya2})/2}\\right)=\\left(\\simplify{{(xa1+xa2)/2}},\\simplify{{(ya1+ya2)/2}}\\right)$.$\\left(\\simplify[basic]{({xa1}+{xa2})/2},\\simplify[basic]{({ya1}+{ya2})/2}\\right)=\\left(\\simplify{{(xa1+xa2)/2}},\\simplify{{(ya1+ya2)/2}}\\right)$.

The easiest way to do this is to convert these numbers in standard form into decimals, do the calculation with decimal forms, then change them back into standard form.

\\[\\begin{align} \\var{B[0]} \\times 10^3 + \\var{B[1]} \\times 10^4 &= \\var{B[0]*10^3} + \\var{B[1]*10^4} \\\\&= \\var{B[0]*10^3 + B[1]*10^4} \\\\&= \\var{(B[0]*10^3 + B[1]*10^4)/10^4} \\times 10^4\\end{align}\\]

\\[\\begin{align} \\var{B[2]} \\times 10^3 - \\var{B[2] - 1.70} \\times 10^2 &= \\var{B[2]*10^3} - \\var{(B[2] - 1.70)*10^2} \\\\&= \\var{(B[2]*10^3) - ((B[2] - 1.70)*10^2)} \\\\&= \\var{(B[2]*10^3 - (B[2] - 1.70)*10^2)/10^3} \\times 10^3\\end{align}\\]

Calculate the following and write the result in standard index form (for example, for $2.01\\times 10^5$ we would write 2.01*10^5 in the gap).

exact value of sin, cos, tan of random(0,90,120,135,150,180,210,225,240,270,300,315,330) degrees

Often we prefer to work with exact values rather than approximations from a calculator. In this question we require you input your answer without decimals and without entering the words sin, cos or tan. For example to input the exact value of $\\sin(60^\\circ)$, which is $\\dfrac{\\sqrt{3}}{2}$, you would input sqrt(3)/2

In particular, the angle of $\\var{theta}^\\circ$ puts the point {textquadrant}.

Since this point falls on an axis and the point is on the unit circle, it is clear that its coordinates are $(\\var{cos({theta}*pi/180)}, \\var{sin({theta}*pi/180)})$. From these, we can conclude that

$\\tan(\\var{theta}^\\circ)=\\dfrac{\\sin(\\var{theta}^\\circ)}{\\cos(\\var{theta}^\\circ)}=\\dfrac{\\var{sin({theta}*pi/180)}}{\\var{cos({theta}*pi/180)}}\\var{if(theta=90 or theta=270, \" which is undefined\",\" = \" + precround(tan(theta*pi/180),0))}.$

The triangle {textquadrant} has the same side lengths as the related triangle in the first quadrant (they are congruent). Therefore, we can recall or use right-angled triangle trigonometry to determine the lengths of the triangle in the first quadrant and then change the signs as needed. Recall:

By drawing the following triangles, we can determine the exact values of $\\sin$, $\\cos$ and $\\tan$ for the angles $30^\\circ$, $45^\\circ$ and $60^\\circ$.

Alternatively, one can memorise the following table:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

	$30^\\circ$	$45^\\circ$	$60^\\circ$

$\\sin$	$\\dfrac{1}{2}$	$\\dfrac{1}{\\sqrt{2}}$	$\\dfrac{\\sqrt{3}}{2}$

$\\cos$	$\\dfrac{\\sqrt{3}}{2}$	$\\dfrac{1}{\\sqrt{2}}$	$\\dfrac{1}{2}$

$\\tan$	$\\dfrac{1}{\\sqrt{3}}$	$1$	$\\sqrt{3}$

From the above, the triangle in the first quadrant tells us that:

$\\sin(\\var{phi}^\\circ)=\\;\\;\\dfrac{1}{2}$$\\sin(\\var{phi}^\\circ)=\\dfrac{1}{\\sqrt{2}}$$\\sin(\\var{phi}^\\circ)=\\dfrac{\\sqrt{3}}{2}$

$\\cos(\\var{phi}^\\circ)=\\dfrac{\\sqrt{3}}{2}$$\\cos(\\var{phi}^\\circ)=\\dfrac{1}{\\sqrt{2}}$$\\cos(\\var{phi}^\\circ)=\\;\\;\\dfrac{1}{2}$

$\\tan(\\var{phi}^\\circ)=\\dfrac{1}{\\sqrt{3}}$$\\tan(\\var{phi}^\\circ)=\\;\\;1$$\\tan(\\var{phi}^\\circ)=\\sqrt{3}$

Since $\\theta=\\var{theta}^\\circ$ puts us {textquadrant}, the $x$-coordinate (the cosine value) is positive,negative, and the $y$-coordinate (the sine value) is positivenegative. That is:

$\\sin(\\var{theta}^\\circ)=\\phantom{-}\\sin(\\var{phi}^\\circ)=\\;\\;\\phantom{-}\\dfrac{1}{2}$$\\sin(\\var{theta}^\\circ)=\\phantom{-}\\sin(\\var{phi}^\\circ)=\\phantom{-}\\dfrac{1}{\\sqrt{2}}$$\\sin(\\var{theta}^\\circ)=\\phantom{-}\\sin(\\var{phi}^\\circ)=\\phantom{-}\\dfrac{\\sqrt{3}}{2}$

$\\cos(\\var{theta}^\\circ)=-\\cos(\\var{phi}^\\circ)=-\\dfrac{\\sqrt{3}}{2}$$\\cos(\\var{theta}^\\circ)=-\\cos(\\var{phi}^\\circ)=-\\dfrac{1}{\\sqrt{2}}$$\\cos(\\var{theta}^\\circ)=-\\cos(\\var{phi}^\\circ)=\\;\\;-\\dfrac{1}{2}$

$\\tan(\\var{theta}^\\circ)=-\\tan(\\var{phi}^\\circ)=-\\dfrac{1}{\\sqrt{3}}$$\\tan(\\var{theta}^\\circ)=-\\tan(\\var{phi}^\\circ)=\\;\\;-1$$\\tan(\\var{theta}^\\circ)=-\\tan(\\var{phi}^\\circ)=-\\sqrt{3}$

$\\sin(\\var{theta}^\\circ)=-\\sin(\\var{phi}^\\circ)=\\;\\;-\\dfrac{1}{2}$$\\sin(\\var{theta}^\\circ)=-\\sin(\\var{phi}^\\circ)=-\\dfrac{1}{\\sqrt{2}}$$\\sin(\\var{theta}^\\circ)=-\\sin(\\var{phi}^\\circ)=-\\dfrac{\\sqrt{3}}{2}$

$\\tan(\\var{theta}^\\circ)=\\phantom{-}\\tan(\\var{phi}^\\circ)=\\phantom{-}\\dfrac{1}{\\sqrt{3}}$$\\tan(\\var{theta}^\\circ)=\\phantom{-}\\tan(\\var{phi}^\\circ)=\\phantom{-}\\;\\;1$$\\tan(\\var{theta}^\\circ)=\\phantom{-}\\tan(\\var{phi}^\\circ)=\\phantom{-}\\sqrt{3}$

$\\cos(\\var{theta}^\\circ)=\\phantom{-}\\cos(\\var{phi}^\\circ)=\\phantom{-}\\dfrac{\\sqrt{3}}{2}$$\\cos(\\var{theta}^\\circ)=\\phantom{-}\\cos(\\var{phi}^\\circ)=\\phantom{-}\\dfrac{1}{\\sqrt{2}}$$\\cos(\\var{theta}^\\circ)=\\phantom{-}\\cos(\\var{phi}^\\circ)=\\;\\;\\phantom{-}\\dfrac{1}{2}$

An alternative approach is to use the mnemonic \"All Stations To Central\" or \"ASTC\" to recall which trig functions are positive in each quadrant (and hence which are negative):

a)

b)