Question

sigma(a+b)tan(A-B/2)=0​

Answer 1

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Answer 2

Answer

Solution of Triangles

Solution of TrianglesThe rule is extremely useful for expressing the sides of a triangle in terms of sines of angle or vice versa as per the requirement of the question. Hence, for the form a/ sin A = b/ sin B = c/ sin C = k, we have,

Solution of TrianglesThe rule is extremely useful for expressing the sides of a triangle in terms of sines of angle or vice versa as per the requirement of the question. Hence, for the form a/ sin A = b/ sin B = c/ sin C = k, we have,∠A = 60°, then b2 + c2 - a2 = bc

Solution of TrianglesThe rule is extremely useful for expressing the sides of a triangle in terms of sines of angle or vice versa as per the requirement of the question. Hence, for the form a/ sin A = b/ sin B = c/ sin C = k, we have,∠A = 60°, then b2 + c2 - a2 = bc∠B = 60°, then a2 + c2 - b2 = ac

Solution of TrianglesThe rule is extremely useful for expressing the sides of a triangle in terms of sines of angle or vice versa as per the requirement of the question. Hence, for the form a/ sin A = b/ sin B = c/ sin C = k, we have,∠A = 60°, then b2 + c2 - a2 = bc∠B = 60°, then a2 + c2 - b2 = acSemi-perimeter of the triangle

Solution of TrianglesThe rule is extremely useful for expressing the sides of a triangle in terms of sines of angle or vice versa as per the requirement of the question. Hence, for the form a/ sin A = b/ sin B = c/ sin C = k, we have,∠A = 60°, then b2 + c2 - a2 = bc∠B = 60°, then a2 + c2 - b2 = acSemi-perimeter of the triangleIf ‘s’ is assumed to be the perimeter of the triangle then s = a + b + c / 2

Solution of TrianglesThe rule is extremely useful for expressing the sides of a triangle in terms of sines of angle or vice versa as per the requirement of the question. Hence, for the form a/ sin A = b/ sin B = c/ sin C = k, we have,∠A = 60°, then b2 + c2 - a2 = bc∠B = 60°, then a2 + c2 - b2 = acSemi-perimeter of the triangleIf ‘s’ is assumed to be the perimeter of the triangle then s = a + b + c / 2Area of a triangle

Solution of TrianglesThe rule is extremely useful for expressing the sides of a triangle in terms of sines of angle or vice versa as per the requirement of the question. Hence, for the form a/ sin A = b/ sin B = c/ sin C = k, we have,∠A = 60°, then b2 + c2 - a2 = bc∠B = 60°, then a2 + c2 - b2 = acSemi-perimeter of the triangleIf ‘s’ is assumed to be the perimeter of the triangle then s = a + b + c / 2Area of a triangleIf Δ is the area of the triangle ABC then

Solution of TrianglesThe rule is extremely useful for expressing the sides of a triangle in terms of sines of angle or vice versa as per the requirement of the question. Hence, for the form a/ sin A = b/ sin B = c/ sin C = k, we have,∠A = 60°, then b2 + c2 - a2 = bc∠B = 60°, then a2 + c2 - b2 = acSemi-perimeter of the triangleIf ‘s’ is assumed to be the perimeter of the triangle then s = a + b + c / 2Area of a triangleIf Δ is the area of the triangle ABC thenΔ = 1/2 bc sin A = 1/2 ca sin B = 1/2 ab sin C

Solution of TrianglesThe rule is extremely useful for expressing the sides of a triangle in terms of sines of angle or vice versa as per the requirement of the question. Hence, for the form a/ sin A = b/ sin B = c/ sin C = k, we have,∠A = 60°, then b2 + c2 - a2 = bc∠B = 60°, then a2 + c2 - b2 = acSemi-perimeter of the triangleIf ‘s’ is assumed to be the perimeter of the triangle then s = a + b + c / 2Area of a triangleIf Δ is the area of the triangle ABC thenΔ = 1/2 bc sin A = 1/2 ca sin B = 1/2 ab sin CΔ = √s(s - a) (s - b) (s - c), this is also called as the Hero’s Formula

Solution of TrianglesThe rule is extremely useful for expressing the sides of a triangle in terms of sines of angle or vice versa as per the requirement of the question. Hence, for the form a/ sin A = b/ sin B = c/ sin C = k, we have,∠A = 60°, then b2 + c2 - a2 = bc∠B = 60°, then a2 + c2 - b2 = acSemi-perimeter of the triangleIf ‘s’ is assumed to be the perimeter of the triangle then s = a + b + c / 2Area of a triangleIf Δ is the area of the triangle ABC thenΔ = 1/2 bc sin A = 1/2 ca sin B = 1/2 ab sin CΔ = √s(s - a) (s - b) (s - c), this is also called as the Hero’s FormulaΔ = (a2 sin B sin C)/ 2 sin (B + C) = (b2 sin C sin A)/ 2 sin (A + C) = (c2 sin A sin B)/ 2 sin (A + B)