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Length of the median, passes through vertex A ot ABC triangle ABC is m. It makes angles alpha and beta with side AB and AC respectively. Prove that,

ⅰ) 2m (sin(alpha)-sin(beta)) = a (sin B - sin C)
ii) 2m sin [½ (alpha-beta)] = (b-c)sin(A/2)

User MadHatter
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1 Answer

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Final answer:

To solve the problem, we apply trigonometric identities and the laws of sines and cosines to express the sides of the triangle in terms of the angles and median length. Sum-to-product identities are particularly useful in handling the expressions involving differences in sines. The proof involves substitution and simplification to demonstrate the validity of the given statements.

Step-by-step explanation:

To prove the given statements regarding the median of a triangle and its relations to the triangle's side lengths and angles, we will use trigonometric identities and properties of triangles.

The length of the median m is given, and it makes angles alpha and beta with sides AB and AC respectively. According to the statement we need to prove:

2m(sin(alpha) - sin(beta)) = a(sin(B) - sin(C))

Using the sine rule which states that:

a/sin(A) = b/sin(B) = c/sin(C),

we can express the sides a, b, and c in terms of the angles and the median length m.

The second statement to be proved is:

2m sin(1/2(alpha - beta)) = (b - c)sin(A/2)

We'll use trigonometric identities, specifically the sum-to-product identities to transform the expressions for the difference in sines into a product involving half-angles, such as:

sin(alpha) - sin(beta) = 2 sin(1/2(alpha - beta)) cos(1/2(alpha + beta))

By substituting and simplifying using the given angles and known identities, we'll show how the statement holds.

User Don F
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