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1.display your name on the monitor screen. 2.read in a distance in km and display it in km and meters eg 47.45km would convert to 47 km , 450 meters write code to solve the following equation x = squareroot of asq - bsq the values of a and b to be float read from the keyboard
What else can I help you with?
Write a c program to solve cos series?
Please do.
How do you use tan in a c program?
double x, y; ... x = tan (y);
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).
Program for sin series in c language?
find the program in c-pgms.blogspot.com
Write a c program Fibonacci series using for loop in java?
Exactly what do you mean by 'C program in Java'
C Program for finding tanx series?
#include#include#include#define pi 3.14void main(){float degree,tanvalue;clrscr();printf("Enter degree:\n");scanf("%f",°ree);tanvalue=tan(degree*(pi/180.0));printf("tan value of %.2f is %.2f",degree,tanvalue);getch();}
How do you solve discrete series by grouping?
You will need to use the distributive law to solve discrete series by grouping. The distributive law is a(b + c) = ab + ac. You will be removing the common factors as you go.
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
Write a C program to print the following series 112 122 . 1n2?
write a program to print the series 1/12+1/22+.........+1/n2 ?
How would you solve the integral of 1 plus tan2x plus tan squared 2x?
Integral of 1 is x Integral of tan(2x) = Integral of [sin(2x)/cos(2x)] =-ln (cos(2x)) /2 Integral of tan^2 (2x) = Integral of sec^2(2x)-1 = tan(2x)/2 - x Combining all, Integral of 1 plus tan(2x) plus tan squared 2x is x-ln(cos(2x))/2 +tan(2x)/2 - x + C = -ln (cos(2x))/2 + tan(2x)/2 + C
What has the author C K Tan written?
C. K. Tan has written: 'Strategic management'
How do you use the basic trigonometry functions?
To find unknown sides or angles of a triangle. For triangle ABC, if C is a right angle and you are using angle A the side a is the opposite of A, side b is the adjacent side of angle A and c is the hypotenuse. Ex: Sin A = a/c, if you know any 2 you can solve for the 3rd. Cos A = b/c, if you know any 2 you can solve for the 3rd. Tan A = a/b, and again, if you know any 2 you can solve for the 3rd.
