Math

posted by Vin Mwale

tan 39°=40/AP, So AP =40/tan. in triangle AQP, angle Q= 60/sin 91°=AP/SinQ. Then bearing of Qfrom P = 30° 180°-(91 angle Q).

  1. Steve

    40/tan makes no sense

Respond to this Question

First Name

Your Answer

Similar Questions

  1. math trigg.

    which triangles do you use COS SIN or TAN?
  2. maths please help

    how do you find the isoperimetric quotient of an n sided figure?
  3. Trig

    Given: cos u = 3/5; 0 < u < pi/2 cos v = 5/13; 3pi/2 < v < 2pi Find: sin (v + u) cos (v - u) tan (v + u) First compute or list the cosine and sine of both u and v. Then use the combination rules sin (v + u) = sin u cos …
  4. calculus

    I need help on finding the local linear approximation of tan 62 degree. i got 1.77859292096 can someone check if i got it right?
  5. Trig

    sin(2 tan-1 7/24) (that is inverse tangent) Draw a 7 24 25 right triangle. Label the angle whose tangent is 7/24. What is its angle?
  6. Pre Cal

    could someone please tell me what the other angle identity of tan is?
  7. Trigonometry

    Find the exact value of tan(a-b) sin a = 4/5, -3pi/2<a<-pi; tan b = -sqrt2, pi/2<b<pi identity used is: tan(a-b)=(tan a-tan b)/1+tan a tan b simplify answer using radicals. (a is alpha, b is beta)
  8. Math

    How does the related acute angle help to determine the trigonometric ratios of angles greater than 90 degrees?
  9. precalculus

    For each of the following determine whether or not it is an identity and prove your result. a. cos(x)sec(x)-sin^2(x)=cos^2(x) b. tan(x+(pi/4))= (tan(x)+1)/(1-tan(x)) c. (cos(x+y))/(cos(x-y))= (1-tan(x)tan(y))/(1+tan(x)tan(y)) d. (tan(x)+sin(x))/(1+cos(x))=tan(x) …
  10. Precalculus

    I am trying to submit this homework in but i guess i'm not doing it in exact values because it is not accepting it. I know i'm supposed to be using half angle formulas but maybe the quadrants are messing me up. Please help! Find sin …

More Similar Questions