in triangle abc, if
sin c= (sin a + sin b )/ ( cos a + cos b )
prove that triangle abc is a right-angle triangle.
In a right triangle, assuming a and b are the acute angles, then a + b = 90 degrees and sin b = cos a, and sin a = cos b. Therefore
sin c = (sin a + sin b/(cos a + cos b)
= (cos a + cos b)/(cos a + cos b) = 1
Therefore c is a right angle. This proves that the relation is true for right triangles, if c is the right angle, but does not prove it is not true for other triangles.
To prove that triangle ABC is a right-angled triangle, we need to use the given equation and some trigonometric identities.
Here's how we can approach the proof:
1. Start with the given equation:
sin C = (sin A + sin B) / (cos A + cos B).
2. Recall the trigonometric identity:
sin^2 θ + cos^2 θ = 1.
3. Square both sides of the given equation:
sin^2 C = (sin A + sin B)^2 / (cos A + cos B)^2.
4. Expand the numerator:
sin^2 C = sin^2 A + 2sin A sin B + sin^2 B.
5. Expand the denominator:
(cos^2 A + 2cos A cos B + cos^2 B).
6. Apply the trigonometric identity from step 2:
sin^2 C = 1 - cos^2 C.
7. Substitute the trigonometric identity into the numerator and denominator:
1 - cos^2 C = cos^2 A + 2cos A cos B + cos^2 B.
8. Rearrange the equation:
cos^2 A + cos^2 B - cos^2 C = 2cos A cos B.
9. Let's prove that the left-hand side (LHS) is equal to the right-hand side (RHS):
LHS:
cos^2 A + cos^2 B - cos^2 C
= cos^2 A + cos^2 B - (1 - sin^2 C) (applying the trigonometric identity from step 2)
= cos^2 A + cos^2 B - 1 + sin^2 C
= sin^2 A + sin^2 B - 1 + sin^2 C (applying the trigonometric identity sin^2 θ = 1 - cos^2 θ)
= sin^2 A + sin^2 B + sin^2 C - 1.
RHS:
2cos A cos B.
Therefore, we have:
sin^2 A + sin^2 B + sin^2 C - 1 = 2cos A cos B.
10. Using the Pythagorean identity:
sin^2 A + sin^2 B + sin^2 C = 1.
11. Substitute the identity into the equation from step 9:
1 - 1 = 2cos A cos B.
12. Simplify:
0 = 2cos A cos B.
13. Since the product of two non-zero numbers is zero, at least one of them must be zero:
cos A = 0 or cos B = 0.
14. If either cos A or cos B is zero, it means one of the angles A or B is 90 degrees (a right angle).
15. Therefore, triangle ABC is proven to be a right-angled triangle.
Hence, we have successfully proved that triangle ABC is a right-angled triangle based on the given equation.