Finding Angles from Side Length in Pre-Calculus












1














I had this problem on a recently college exam. I had no idea how to do it, and lost all my points for it. I assumed it would have something to do with finding the correct values for the sides and angles of a triangle given the sine and cosine rules, but the problem didn't seem to give enough space to solve them with that.



On my exam my professor wrote the tangent addition identity:



$$tan(A + B) = frac{tan(A) + tan(B)}{1 - tan(A) cdot tan(B)}$$



But I'm not sure what relevance this has.



If C = A + B, find tan(C)
enter image description here










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  • Are the two angles on the top triangles right angles?
    – KM101
    Nov 21 at 20:45










  • @KM101 Yes, that's correct.
    – LuminousNutria
    Nov 21 at 20:46










  • There is no need for anything really. You’re given the opposite and adjacent sides to $angle A$ and $angle B$, so you can easily find $tan A$ and $tan B$. From there, you use the identity.
    – KM101
    Nov 21 at 20:53


















1














I had this problem on a recently college exam. I had no idea how to do it, and lost all my points for it. I assumed it would have something to do with finding the correct values for the sides and angles of a triangle given the sine and cosine rules, but the problem didn't seem to give enough space to solve them with that.



On my exam my professor wrote the tangent addition identity:



$$tan(A + B) = frac{tan(A) + tan(B)}{1 - tan(A) cdot tan(B)}$$



But I'm not sure what relevance this has.



If C = A + B, find tan(C)
enter image description here










share|cite|improve this question






















  • Are the two angles on the top triangles right angles?
    – KM101
    Nov 21 at 20:45










  • @KM101 Yes, that's correct.
    – LuminousNutria
    Nov 21 at 20:46










  • There is no need for anything really. You’re given the opposite and adjacent sides to $angle A$ and $angle B$, so you can easily find $tan A$ and $tan B$. From there, you use the identity.
    – KM101
    Nov 21 at 20:53
















1












1








1







I had this problem on a recently college exam. I had no idea how to do it, and lost all my points for it. I assumed it would have something to do with finding the correct values for the sides and angles of a triangle given the sine and cosine rules, but the problem didn't seem to give enough space to solve them with that.



On my exam my professor wrote the tangent addition identity:



$$tan(A + B) = frac{tan(A) + tan(B)}{1 - tan(A) cdot tan(B)}$$



But I'm not sure what relevance this has.



If C = A + B, find tan(C)
enter image description here










share|cite|improve this question













I had this problem on a recently college exam. I had no idea how to do it, and lost all my points for it. I assumed it would have something to do with finding the correct values for the sides and angles of a triangle given the sine and cosine rules, but the problem didn't seem to give enough space to solve them with that.



On my exam my professor wrote the tangent addition identity:



$$tan(A + B) = frac{tan(A) + tan(B)}{1 - tan(A) cdot tan(B)}$$



But I'm not sure what relevance this has.



If C = A + B, find tan(C)
enter image description here







algebra-precalculus trigonometry triangle






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share|cite|improve this question











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asked Nov 21 at 20:43









LuminousNutria

1709




1709












  • Are the two angles on the top triangles right angles?
    – KM101
    Nov 21 at 20:45










  • @KM101 Yes, that's correct.
    – LuminousNutria
    Nov 21 at 20:46










  • There is no need for anything really. You’re given the opposite and adjacent sides to $angle A$ and $angle B$, so you can easily find $tan A$ and $tan B$. From there, you use the identity.
    – KM101
    Nov 21 at 20:53




















  • Are the two angles on the top triangles right angles?
    – KM101
    Nov 21 at 20:45










  • @KM101 Yes, that's correct.
    – LuminousNutria
    Nov 21 at 20:46










  • There is no need for anything really. You’re given the opposite and adjacent sides to $angle A$ and $angle B$, so you can easily find $tan A$ and $tan B$. From there, you use the identity.
    – KM101
    Nov 21 at 20:53


















Are the two angles on the top triangles right angles?
– KM101
Nov 21 at 20:45




Are the two angles on the top triangles right angles?
– KM101
Nov 21 at 20:45












@KM101 Yes, that's correct.
– LuminousNutria
Nov 21 at 20:46




@KM101 Yes, that's correct.
– LuminousNutria
Nov 21 at 20:46












There is no need for anything really. You’re given the opposite and adjacent sides to $angle A$ and $angle B$, so you can easily find $tan A$ and $tan B$. From there, you use the identity.
– KM101
Nov 21 at 20:53






There is no need for anything really. You’re given the opposite and adjacent sides to $angle A$ and $angle B$, so you can easily find $tan A$ and $tan B$. From there, you use the identity.
– KM101
Nov 21 at 20:53












2 Answers
2






active

oldest

votes


















1














Hint: $$tan A = frac{1}{2}$$



$$tan B = frac{3}{4}$$



Now, apply $$tan(A+B) = frac{tan A+tan B}{1-tan Atan B}$$






share|cite|improve this answer





















  • How do you know the value of $tan A$ and $tan B$ here? How do you know that the botton side of those triangles is the hypotenuse?
    – LuminousNutria
    Nov 21 at 20:54












  • The bottom side IS the hypotenuse, so the sides given by the question are the opposite and adjacent sides.
    – KM101
    Nov 21 at 20:55












  • Sorry, I just edited my post. How do you know which side is the hypotenuse?
    – LuminousNutria
    Nov 21 at 20:56






  • 2




    It seems to be implied by the “squar-ish” angle marks.
    – KM101
    Nov 21 at 20:58












  • Oh, right. Since the sum of the angles of a triangle is 180 degrees, and 90 degrees is half of that, the other two angles must be smaller. Since the other angles are smaller, the sides opposite them must also be smaller. Therefore the side opposite a 90 degree angle in a triangle is always the hypotenuse. Thanks!
    – LuminousNutria
    Nov 21 at 21:00





















0














$$tan(C) =tan(A+B) = frac{1/2 + 3/4}{1 - 3/8} = frac{5/8}{5/8} = 1$$






share|cite|improve this answer























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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1














    Hint: $$tan A = frac{1}{2}$$



    $$tan B = frac{3}{4}$$



    Now, apply $$tan(A+B) = frac{tan A+tan B}{1-tan Atan B}$$






    share|cite|improve this answer





















    • How do you know the value of $tan A$ and $tan B$ here? How do you know that the botton side of those triangles is the hypotenuse?
      – LuminousNutria
      Nov 21 at 20:54












    • The bottom side IS the hypotenuse, so the sides given by the question are the opposite and adjacent sides.
      – KM101
      Nov 21 at 20:55












    • Sorry, I just edited my post. How do you know which side is the hypotenuse?
      – LuminousNutria
      Nov 21 at 20:56






    • 2




      It seems to be implied by the “squar-ish” angle marks.
      – KM101
      Nov 21 at 20:58












    • Oh, right. Since the sum of the angles of a triangle is 180 degrees, and 90 degrees is half of that, the other two angles must be smaller. Since the other angles are smaller, the sides opposite them must also be smaller. Therefore the side opposite a 90 degree angle in a triangle is always the hypotenuse. Thanks!
      – LuminousNutria
      Nov 21 at 21:00


















    1














    Hint: $$tan A = frac{1}{2}$$



    $$tan B = frac{3}{4}$$



    Now, apply $$tan(A+B) = frac{tan A+tan B}{1-tan Atan B}$$






    share|cite|improve this answer





















    • How do you know the value of $tan A$ and $tan B$ here? How do you know that the botton side of those triangles is the hypotenuse?
      – LuminousNutria
      Nov 21 at 20:54












    • The bottom side IS the hypotenuse, so the sides given by the question are the opposite and adjacent sides.
      – KM101
      Nov 21 at 20:55












    • Sorry, I just edited my post. How do you know which side is the hypotenuse?
      – LuminousNutria
      Nov 21 at 20:56






    • 2




      It seems to be implied by the “squar-ish” angle marks.
      – KM101
      Nov 21 at 20:58












    • Oh, right. Since the sum of the angles of a triangle is 180 degrees, and 90 degrees is half of that, the other two angles must be smaller. Since the other angles are smaller, the sides opposite them must also be smaller. Therefore the side opposite a 90 degree angle in a triangle is always the hypotenuse. Thanks!
      – LuminousNutria
      Nov 21 at 21:00
















    1












    1








    1






    Hint: $$tan A = frac{1}{2}$$



    $$tan B = frac{3}{4}$$



    Now, apply $$tan(A+B) = frac{tan A+tan B}{1-tan Atan B}$$






    share|cite|improve this answer












    Hint: $$tan A = frac{1}{2}$$



    $$tan B = frac{3}{4}$$



    Now, apply $$tan(A+B) = frac{tan A+tan B}{1-tan Atan B}$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 21 at 20:52









    KM101

    3,970417




    3,970417












    • How do you know the value of $tan A$ and $tan B$ here? How do you know that the botton side of those triangles is the hypotenuse?
      – LuminousNutria
      Nov 21 at 20:54












    • The bottom side IS the hypotenuse, so the sides given by the question are the opposite and adjacent sides.
      – KM101
      Nov 21 at 20:55












    • Sorry, I just edited my post. How do you know which side is the hypotenuse?
      – LuminousNutria
      Nov 21 at 20:56






    • 2




      It seems to be implied by the “squar-ish” angle marks.
      – KM101
      Nov 21 at 20:58












    • Oh, right. Since the sum of the angles of a triangle is 180 degrees, and 90 degrees is half of that, the other two angles must be smaller. Since the other angles are smaller, the sides opposite them must also be smaller. Therefore the side opposite a 90 degree angle in a triangle is always the hypotenuse. Thanks!
      – LuminousNutria
      Nov 21 at 21:00




















    • How do you know the value of $tan A$ and $tan B$ here? How do you know that the botton side of those triangles is the hypotenuse?
      – LuminousNutria
      Nov 21 at 20:54












    • The bottom side IS the hypotenuse, so the sides given by the question are the opposite and adjacent sides.
      – KM101
      Nov 21 at 20:55












    • Sorry, I just edited my post. How do you know which side is the hypotenuse?
      – LuminousNutria
      Nov 21 at 20:56






    • 2




      It seems to be implied by the “squar-ish” angle marks.
      – KM101
      Nov 21 at 20:58












    • Oh, right. Since the sum of the angles of a triangle is 180 degrees, and 90 degrees is half of that, the other two angles must be smaller. Since the other angles are smaller, the sides opposite them must also be smaller. Therefore the side opposite a 90 degree angle in a triangle is always the hypotenuse. Thanks!
      – LuminousNutria
      Nov 21 at 21:00


















    How do you know the value of $tan A$ and $tan B$ here? How do you know that the botton side of those triangles is the hypotenuse?
    – LuminousNutria
    Nov 21 at 20:54






    How do you know the value of $tan A$ and $tan B$ here? How do you know that the botton side of those triangles is the hypotenuse?
    – LuminousNutria
    Nov 21 at 20:54














    The bottom side IS the hypotenuse, so the sides given by the question are the opposite and adjacent sides.
    – KM101
    Nov 21 at 20:55






    The bottom side IS the hypotenuse, so the sides given by the question are the opposite and adjacent sides.
    – KM101
    Nov 21 at 20:55














    Sorry, I just edited my post. How do you know which side is the hypotenuse?
    – LuminousNutria
    Nov 21 at 20:56




    Sorry, I just edited my post. How do you know which side is the hypotenuse?
    – LuminousNutria
    Nov 21 at 20:56




    2




    2




    It seems to be implied by the “squar-ish” angle marks.
    – KM101
    Nov 21 at 20:58






    It seems to be implied by the “squar-ish” angle marks.
    – KM101
    Nov 21 at 20:58














    Oh, right. Since the sum of the angles of a triangle is 180 degrees, and 90 degrees is half of that, the other two angles must be smaller. Since the other angles are smaller, the sides opposite them must also be smaller. Therefore the side opposite a 90 degree angle in a triangle is always the hypotenuse. Thanks!
    – LuminousNutria
    Nov 21 at 21:00






    Oh, right. Since the sum of the angles of a triangle is 180 degrees, and 90 degrees is half of that, the other two angles must be smaller. Since the other angles are smaller, the sides opposite them must also be smaller. Therefore the side opposite a 90 degree angle in a triangle is always the hypotenuse. Thanks!
    – LuminousNutria
    Nov 21 at 21:00













    0














    $$tan(C) =tan(A+B) = frac{1/2 + 3/4}{1 - 3/8} = frac{5/8}{5/8} = 1$$






    share|cite|improve this answer




























      0














      $$tan(C) =tan(A+B) = frac{1/2 + 3/4}{1 - 3/8} = frac{5/8}{5/8} = 1$$






      share|cite|improve this answer


























        0












        0








        0






        $$tan(C) =tan(A+B) = frac{1/2 + 3/4}{1 - 3/8} = frac{5/8}{5/8} = 1$$






        share|cite|improve this answer














        $$tan(C) =tan(A+B) = frac{1/2 + 3/4}{1 - 3/8} = frac{5/8}{5/8} = 1$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 21 at 22:01









        Davide Giraudo

        125k16150259




        125k16150259










        answered Nov 21 at 20:50









        WhatToDo

        25116




        25116






























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