Expert: Paul Klarreich - 2/6/2011

Question
deriving sine and cosine rules?
I need help with this worded trig problem
Draw a non-right angled triangle. Mark the Vertices A,B,and C. Mark the sides a,b,and c opposite angles A,B, and C. Turn your drawing if necessary so that the triangle is sitting on a base with an acute angle at the top. Draw in an altitude to cut the triangle into two right-angled triangles. Using each right-angled triangle in turn, write down an expression for the height of the altitude. Put these two expressions together and rearrange them to get one part of the sine rule. Use another altitude to get the other equality. Check an obtuse angled triangle case in which some of the altitudes meet the base out-side the triangle.
The cosine rule with a^2, b^2 and c^2 is a general case of pythagoras' theorem. Draw a triangle, label the sides and angles, draw in an altitude (say height h). now write down an expression of Pythagoras' theorem for each right-angle in turn- if the base is c you will meed to label its two parts x and c-x

For each of these two expressions make h^2 the subject on the left-hand side. Now each of the right hand sides of both expressions is equal to h^2 and hence equal to each other.
Writing this equality down gives you one form of the cosine rule. It can be written with cos A, Cos B,cos C, a^2, b^2 or c^2, as the subject of the formula depending on which two right angled triangles you start with.

Any good in depth explanation will be a life saver
Thanks.

Answer
Questioner: vidhi
Country: United Kingdom
Category: Advanced Math
Private: Yes
Subject: trig
Question: deriving sine and cosine rules?
I need help with this worded trig problem
Draw triangle ABC. Mark the sides a,b,and c opposite angles A,B, and C. Turn your drawing if necessary so that the triangle is sitting on a base with an acute angle at the top. Draw in an altitude to cut the triangle into two right-angled triangles. Using each right-angled triangle in turn, write down an expression for the height of the altitude. Put these two expressions together and rearrange them to get one part of the sine rule. Use another altitude to get the other equality. Check an obtuse angled triangle case in which some of the altitudes meet the base out-side the triangle.
The cosine rule with a^2, b^2 and c^2 is a general case of pythagoras' theorem. Draw a triangle, label the sides and angles, draw in an altitude (say height h). now write down an expression of Pythagoras' theorem for each right-angle in turn- if the base is c you will meed to label its two parts x and c-x

For each of these two expressions make h^2 the subject on the left-hand side. Now each of the right hand sides of both expressions is equal to h^2 and hence equal to each other.
Writing this equality down gives you one form of the cosine rule. It can be written with cos A, Cos B,cos C, a^2, b^2 or c^2, as the subject of the formula depending on which two right angled triangles you start with.

Any good in depth explanation will be a life saver
Thanks.
.................................................
Hi, Vaidhya,
(I assume that is your name -- that is what I call my friend and bridge partner.)

Expl: This is nothing more than a fairly standard derivation of the Law of Sines and Law of Cosines.  Just do exactly what the words say (condensed below.)

1.Draw triangle ABC.
2.Mark a,b,and c opposite A,B, and C.
3.Put acute angle A at the top.
4.Draw an altitude from A to point D, on BC.
5.Call AD = h.
6. (sorry -- there is no 6)
7a. In triangle ABD, write h in terms of B and c. (Hint: h = c sin B)   *********
7b. In triangle ACD, write h in terms of C and b.                       *********
                      
8.Put these two expressions together and rearrange them to get one part of the sine rule.

Use an altitude from B the same way to get the other equality.

Do it all over with an obtuse-angled triangle for practice.
......................................

Start over.
Do 1-5.
6. Label the parts  BD = x and  DC =  a-x

7A. Apply Pythagoras' theorem in triangle ABD
7B. Apply Pythagoras' theorem in triangle ACD.

   x^2 + h^2 = c^2
(a-x)^2 + h^2 = b^2

8.Put them together to get an expression for b^2.

b^2 = (a-x)^2 + h^2

b^2 = a^2 + x^2 - 2ax + h^2

But x^2 + h^2 = c^2, so:

b^2 = a^2 + c^2 - 2ax

Finally express x in terms of angle B and c.

Over to you, now.