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What is the birth name of Yce Tan?
Yce Tan's birth name is Rycelonia Tan.
What is the birth name of Chinkee Tan?
Chinkee Tan's birth name is Ferdinand Tan.
What is the birth name of Teck Tan?
Teck Tan's birth name is Tan Chor Teck.
What is the tan in black and tan?
Guinness over Bass
What is the name of Arvic Tan's girlfriend?
Mika Tan :)
Find the cot of a 32 degree angle?
cot 32° = 1/(tan 32°) = 1/(0.6249) = 1.6003
How old is Laurentia Tan?
Laurentia Tan is 32 years old (birthdate: April 24, 1979).
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
The angle of depression from a lighthouse to a ship at sea is 32 degrees. Find the distance the ship is from lighthouse if the height of the lighthouse is 20 meters to the nearest meter?
The alternate angle is 32 degrees and so 20/tan(32) = 32 meters
What is the maximum acceleration a car can undergo if the coefficient of static friction between the tires and the ground is 0.78?
coefficient of friction = 0.8 tan theta use your calculator : 2nd tan ( 0.8) = 38.66 degrees
Tan 9 plus tan 81 -tan 27-tan 63?
tan(9) + tan(81) - tan(27) - tan(63) = 4
What is a antonym of pale?
Tan Tan
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).
How do you integrate one over sqrt of x squared minus nine between limits five and ten?
5∫10[1/√(x2 - 9)] dx = 5∫10[dx/√(x2 - 32)] Let x = 3 sec θ, where 0 < θ < π/2 or π < θ < 3π/2. Then, dx = 3 sec θ tan θ dθ and √(x2 - 9) = √[(3 sec θ)2 - 9] = √[9(sec2 θ - 1)] = √(9 tan2 θ) = 3 tan θ 5∫10[dx/√(x2 - 32)] = 5∫10[(√3 sec θ tan θ)/(3 tan θ)] dθ = 5∫10 sec θ dθ = ln |sec θ + tan θ|5|10 = ln sec 10 + tan 10 - sec 5 - tan 5 = -2.298
When was Tan Cerca...Tan Lejos created?
Tan Cerca...Tan Lejos was created in 1975.
What is the airport code for Tan Tan Airport?
The airport code for Tan Tan Airport is TTA.
What is the exact trigonometric function value of cot 15 degrees?
cot(15)=1/tan(15) Let us find tan(15) tan(15)=tan(45-30) tan(a-b) = (tan(a)-tan(b))/(1+tan(a)tan(b)) tan(45-30)= (tan(45)-tan(30))/(1+tan(45)tan(30)) substitute tan(45)=1 and tan(30)=1/√3 into the equation. tan(45-30) = (1- 1/√3) / (1+1/√3) =(√3-1)/(√3+1) The exact value of cot(15) is the reciprocal of the above which is: (√3+1) /(√3-1)
