yes, israel is very hot and you would get a very chocolate brown tan!
It could be a purebred, but it is not desirable for a show dog.
Some names for different shades of brown are .... tan, beige, fawn, chocolate etc
Yes. First you go to the ocean and stay there: In one hour you will get a tan In Two hours the tan will be a bit darker In Three hours will give the tan the colour of a nut In Four hours the tan will be a chocolate colour In Five hours the tan will be dark chocolate colour Addendum: The 'hours' spent is in real-time, an hour on the beach in-game will correspond to an hour in the real world.
I am not saying that a pitbull can't be black and tan but I never seen these color in a pitbull. I have seen tan dogs or black dogs. What I have seen is bullies that are tan and black. If you want a real pitbull, go for the body characteristics instead of the color.
Colors such as tan or chocolate are very in right now.
It depends on your skin type. To prevent your skin from tanning you should use anti sun tan lotion. But as an alternative you can also use chocolate milk as well.
tan(9) + tan(81) - tan(27) - tan(63) = 4
Silk?
Probably the best color for light tan skin is something lighter than you're original hair color. Like for example, a chocolate brown hair color wouldn't look bad... don't go for blond
Tan Tan
This may not be the most efficient method but ... Let the three angle be A, B and C. Then note that A + B + C = 20+32+38 = 90 so that C = 90-A+B. Therefore, sin(C) = sin[(90-(A+B) = cos(A+B) and cos(C) = cos[(90-(A+B) = sin(A+B). So that tan(C) = sin(C)/cos(C) = cos(A+B) / sin(A+B) = cot(A+B) Now, tan(A+B) = [tan(A)+tan(B)] / [1- tan(A)*tan(B)] so cot(A+B) = [1- tan(A)*tan(B)] / [tan(A)+tan(B)] The given expressin is tan(A)*tan(B) + tan(B)*tan(C) + tan(C)*tan(A) = tan(A)*tan(B) + [tan(B) + tan(A)]*cot(A+B) substituting for cot(A+B) gives = tan(A)*tan(B) + [tan(B) + tan(A)]*[1- tan(A)*tan(B)]/[tan(A)+tan(B)] cancelling [tan(B) + tan(A)] and [tan(A) + tan(B)], which are equal, in the second expression. = tan(A)*tan(B) + [1- tan(A)*tan(B)] = 1