InnovativeProgrammer

$${\mathtt{4}}{\mathtt{\,\times\,}}\left({\mathtt{tan3}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{sin5}}}^{{\mathtt{0.5}}}\right) = {\mathtt{1.390\: \!517\: \!175\: \!974\: \!257\: \!7}}$$

Same as atan.

Press shift and click on tan, sin or cos

I don't know what tital is :P
If you mean total volume them it's as simple as:

$${\mathtt{120}}{\mathtt{\,\times\,}}{\mathtt{120}}{\mathtt{\,\times\,}}{\mathtt{80}} = {\mathtt{1\,152\,000}}$$

$${\mathtt{1\,152\,000}}{\mathtt{\,\times\,}}{\mathtt{10}} = {\mathtt{11\,520\,000}}$$cm3

11 520 000cm3 = 11 520 dm3 = 11,52 m3

It's the same as asin.

Press shift and and click sin

Not quite sure about this so if anybody could confirm please do so.

The opposite of tan is $${atan}\left($$ or tan-1

So the calculation is atan(0.217)=12.24

This question is in Danish but luckly I'm a Dane so I will answer in both english and Danish

Danish:

Meget enkelt. Du har en hest og du siger minus hest = ingenting

Det er egentlig ligemeget at hest = ged fordi du har ingen heste

Hest-Hest=Ged-Ged

Hvem spørger om sådan et spørgsmål?

English:

The question he is asking is: What is horse-hose if horse=goat

Simple enough. Horse-Horse = nothing

So it does not really matter but Horse-Horse=Goat-Goat

I have way too much time on my hands.

Enjoy the Danish

If I understood your question correctly the answer should be like this:

1 1/16 is the same as 17/16 so

$${\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{17}} = {\mathtt{68}}$$, $${\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{16}} = {\mathtt{80}}$$

68/80 = 17/20. Right?.

Sorry for the late answer but here goes:

You simply select sin, cos or tan. Then it should say something like $${tan}\left($$

Type the amount of degrees the angle is and close the parenthesis: $$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{50}}^\circ\right)}$$

So let's say I wanted to use tan to figure out a in an right-angled triangle. I have Angle A and side a

The formula is $${\frac{{\mathtt{a}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{A}}\right)}}}$$, so I go: $${\frac{{\mathtt{5}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{50}}^\circ\right)}}} = {\mathtt{4.195\: \!498\: \!155\: \!887\: \!139\: \!2}}$$

Hope this helped even though it probably didn't...