tan(9) + tan(81) - tan(27) - tan(63) = 4
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).
The answer is 74, 100% of 74 = 74.
74% of 40= 74% * 40= 0.74 * 40= 29.6
82% of 74 = 82% * 74 = 0.82 * 74 = 60.68
Settantaquattro is an Italian equivalent of the English word "seventy-four (74)." The numbers represents the combination of settanta ("seventy," "70") and quattro ("four," "4"). The pronunciation will be "set-TAN-ta-KWAT-tro" in Italian.
tan(9) + tan(81) - tan(27) - tan(63) = 4
Tan Tan
Oh honey, you're throwing some trigonometry at me? Alright, buckle up. The sum of tan20tan32 plus tan32tan38 plus tan38tan20 is equal to 1. Just plug in those values and watch the magic happen. Math can be sassy too, you know.
Add a percent sign (74%). 74 = 7400% .74 = 74%
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).
The answer is 74, 100% of 74 = 74.
74
Tan Cerca...Tan Lejos was created in 1975.
The airport code for Tan Tan Airport is TTA.
74% of 40= 74% * 40= 0.74 * 40= 29.6
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)