0
To find the angle ( a ) for which ( \tan a = 0.42 ), you can use the inverse tangent function (also known as arctangent). This can be calculated using a calculator or software that provides trigonometric functions. The result will give you ( a ) in radians or degrees, depending on your calculator's settings. Specifically, ( a \approx 22.62^\circ ) or ( a \approx 0.394 ) radians.
Still curious? Ask our experts.
Chat with our AI personalities
Judy
Rene
DevinAdd your answer:
Earn +20 pts
Tan 9 plus tan 81 -tan 27-tan 63?
tan(9) + tan(81) - tan(27) - tan(63) = 4
What is .042 in written form?
Forty-two thousandths
What is 762 plus 80?
762+80=842
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).
What is .042 as a fraction?
21/50
