hectictar

Maybe you should show the working of these stages:

  1  Find the midpoint of AB, and thus, the coordinates of point D.

  2. Find the midpoint of AC, and thus, the coordinates of point E.

  3. Calculate the slope of DE.

  4. Calculate the slope of BC.

Show how you find each of those things.

Then, at the end, add the same concluding sentence which you have written.

One unit on the time axis is  1/8  of an hour, and one unit on the velocity axis is 10 km/h.

1/8 h * 10 km/h  =  5/4 km 

Also, there are 100 squares underneath the blue line, and each square  =  5/4 km

100 * 5/4 km  =  125 km

If the slope of line DE  =  the slope of line BC, then DE  is parallel to  BC.

D  is the midpoint of AB, so    D   =   \((\frac{4+2}{2},\frac{6-2}{2})\)   =   (3, 2)

E  is the midpoint of  AC, so   E   =   \((\frac{4-2}{2},\frac{6-4}{2})\)   =   (1, 1)

slope of  DE  =  \(\frac{y_2-y_1}{x_2-x_1}\,=\,\frac{2-1}{3-1}\,=\,\frac{1}{2}\)

slope of  BC  =  \(\frac{y_2-y_1}{x_2-x_1}\,=\,\frac{-4+2}{-2-2}\,=\,\frac{-2}{-4}\,=\,\frac{1}{2}\)

Since the slope of  DE  and the slope of  BC  both equal  \(\frac12\) ,  DE  is parallel to  BC.

Adding  \(\frac{\pi}{2}\)  flips the tangent and negates it. If     \(\tan\theta\,=\,\frac{a}{b}\)     then    \(\tan(\theta+\frac{\pi}{2})\,=\,-\frac{b}{a}\)    

Let  CO  =  x

△COB  is a  30° - 60° - 90°  triangle. So     BC  =  2x

△COA  is a  45° - 45° - 90°  triangle. So     AC  =  √2 x

And  x  is positive, so     √2 x  <  2x .  So     AC  <  BC

AC  =   √3

√2 x  =  √3

x  =  √3 / √2       Multiply the numerator and denominator by  √2

x  =  √6 / 2

BC  =  2x

BC  =  2( √6 / 2 )

BC  =  √6

By the Pythagorean theorem,

AC² + BC²  =  AB²

(√3)²  +  (√6)²  =  AB²

3  +  6  =  AB²

9  =  AB²           AB is a length, so take the positive sqrt of both sides.

3  =  AB

Yep they're the same!

Do you know how to do this:   log₁₀(12)    ?

I don't know either, but I guess it wouldn't hurt. Maybe better to be safe than sorry!

(a)

\(\sin^2\theta+\cos^2\theta\,=\,1 \)       by the Pythagorean Identity.

\((\frac13)^2+\cos^2\theta\,=\,1\)        because we are given that  \(\sin\theta\,=\,\frac13\)

\(\frac19+\cos^2\theta\,=\,1\\~\\ \cos^2\theta\,=\,1-\frac19\\~\\ \cos^2\theta\,=\,\frac89\\~\\ \cos\theta\,=\,\pm\sqrt{\frac89}\\~\\ \cos\theta\,=\,\pm\frac{2\sqrt2}{3} \)

Since  θ  is in Quadrant II,  cos θ  must be negative. So   \(\cos\theta\,=\,-\frac{2\sqrt2}{3}\)

(b)

\(\sin(\theta+\frac{\pi}{6})\,=\,\sin\theta\,\cos\frac{\pi}{6}+\cos\theta\,\sin\frac{\pi}{6}\)     by the angle sum formula for sin

\(\sin(\theta+\frac{\pi}{6})\,=\,(\frac13)(\frac{\sqrt3}{2})+(-\frac{2\sqrt2}{3})(\frac12)\\~\\ \sin(\theta+\frac{\pi}{6})\,=\,\frac{\sqrt3}{6}-\frac{2\sqrt2}{6}\\~\\ \sin(\theta+\frac{\pi}{6})\,=\,\frac{\sqrt3-2\sqrt2}{6}\)

(c)

\(\cos(\theta-\frac{\pi}{3})\,=\,\cos\theta\,\cos\frac{\pi}{3}+\sin\theta\,\sin\frac{\pi}{3}\)     by the angle difference formula for cos

Now plug in the values for  cos θ,  cos(pi/3) ,  sin θ , and  sin(pi/3)  and simplify. Can you finish this one?

(d)

To use the angle sum formula for tan, let's first find  tan θ.

\(\tan\theta\,=\,\frac{\sin\theta}{\cos\theta}\,=\,\frac{(\frac13)}{(-\frac{2\sqrt2}{3})}\,=\,-\frac{1}{2\sqrt2}\,=\,-\frac{\sqrt2}{4}\)

\(\tan(\theta+\frac{\pi}{4})\,=\,\frac{\tan\theta+\tan\frac{\pi}{4}}{1-\tan\theta\,\tan\frac{\pi}{4}}\)     by the angle sum formula for tan

\(\tan(\theta+\frac{\pi}{4})\,=\,\frac{(-\frac{\sqrt2}{4})+(1)}{1-(-\frac{\sqrt2}{4})(1)}\\~\\ \tan(\theta+\frac{\pi}{4})\,=\,\frac{4-\sqrt2}{4+\sqrt2}\)