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the sun.
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∙ 15y ago
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How can you become tan in less time?
Tanning salon.
What layer of skin causes a tan to become leathery?
scleroderma
How deep can a dead body submerge in water?
It will become white/tan and will be rotten if its not in the water.
What is a quick way to become black and tan?
Pale ale and Guinness are used to make a Black and Tan alcoholic beverage. It is said that pale ale may be used to substitute ale when making the perfect Black and Tan alcoholic beverage.
How do you get the frickn tan cheat?
The only way to get the tan cheat is to transfer wardrobes with someone who has it. So, you have to become friends with them, or simply ask for the tan cheat, and they can most likely do that transfer for you. Of course, most people with it are rude as hell, but its worth a shot. <3
Shihpoo puppy has a brown coat but mother is a white poodle and father is a tan shitztu. Will the puppy change color or stay brown?
I think it's coat will become lighter.It's obvious when you think about genetics. The mother is white, and the father is tan, which equals when he is older that he will be tan.
Tan 9 plus tan 81 -tan 27-tan 63?
tan(9) + tan(81) - tan(27) - tan(63) = 4
What qualifications do you need to become a beatician?
none, just good people skills and lots of fake tan.
In which Family Guy episode does Peter become a cop?
The Tan Aquatic with Steve Zissou
What is a antonym of pale?
Tan Tan
What is tan20tan32 plus tan32tan38 plus tan38tan20?
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
If for a triangle abc tan a-b plus tan b-c plus tan c-a equals 0 then what can you say about the triangle?
tan (A-B) + tan (B-C) + tan (C-A)=0 tan (A-B) + tan (B-C) - tan (A-C)=0 tan (A-B) + tan (B-C) = tan (A-C) (A-B) + (B-C) = A-C So we can solve tan (A-B) + tan (B-C) = tan (A-C) by first solving tan x + tan y = tan (x+y) and then substituting x = A-B and y = B-C. tan (x+y) = (tan x + tan y)/(1 - tan x tan y) So tan x + tan y = (tan x + tan y)/(1 - tan x tan y) (tan x + tan y)tan x tan y = 0 So, tan x = 0 or tan y = 0 or tan x = - tan y tan(A-B) = 0 or tan(B-C) = 0 or tan(A-B) = - tan(B-C) tan(A-B) = 0 or tan(B-C) = 0 or tan(A-B) = tan(C-B) A, B and C are all angles of a triangle, so are all in the range (0, pi). So A-B and B-C are in the range (- pi, pi). At this point I sketched a graph of y = tan x (- pi < x < pi) By inspection I can see that: A-B = 0 or B-C = 0 or A-B = C-B or A-B = C-B +/- pi A = B or B = C or A = C or A = C +/- pi But A and C are both in the range (0, pi) so A = C +/- pi has no solution So A = B or B = C or A = C A triangle ABC has the property that tan (A-B) + tan (B-C) + tan (C-A)=0 if and only if it is isosceles (or equilateral).
