This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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a) Proof: Let ABC be a triangle such that B C. We need to prove that the sides opposite to these angles are congruent, i.e., AC AB. Step 1: Draw the angle bisector of A. Draw a line segment AD from vertex A to side BC such that AD bisects A. This means BAD CAD. Step 2: Consider the two triangles formed. Now we have two triangles: ABD and ACD. Step 3: Apply the Angle-Angle-Side (AAS) congruence criterion. In ABD and ACD: B C (Given) BAD CAD (By construction, AD is the angle bisector of A) AD AD (Common side to both triangles) Therefore, by the AAS congruence criterion, ABD ACD. Step 4: Conclude based on CPCTC. Since the two triangles are congruent, their corresponding parts are congruent (CPCTC). Thus, the side AB in ABD corresponds to the side AC in ACD. Hence, AB AC. This proves that if two angles of a triangle are congruent, then the sides opposite to them are also congruent. b) Given the two polynomials: P(x) = x^2 - 11x + 24 Q(x) = x^2 - 6x + 9 And their HCF is H(x) = (x-3). We need to find the LCM. Step 1: Factorize the first polynomial P(x). We look for two numbers that multiply to 24 and add up to -11. These numbers are -3 and -8. P(x) = x^2 - 11x + 24 = (x-3)(x-8) Step 2: Factorize the second polynomial Q(x). The polynomial x^2 - 6x + 9 is a perfect square trinomial, which can be factored as (x-3)^2. Q(x) = x^2 - 6x + 9 = (x-3)(x-3) = (x-3)^2 Step 3: Use the relationship between HCF, LCM, and the polynomials. The product of two polynomials is equal to the product of their HCF and LCM: P(x) × Q(x) = H(x) × L(x) Where L(x) is the LCM. We can rearrange this to find L(x): L(x) = (P(x) × Q(x))/(H(x)) Step 4: Substitute the factored forms and the given HCF into the formula. L(x) = (((x-3)(x-8)) × ((x-3)^2))/((x-3)) Step 5: Simplify the expression to find the LCM. Cancel one (x-3) term from the numerator and the denominator. L(x) = (x-8)(x-3)^2 We can expand this expression: L(x) = (x-8)(x^2 - 6x + 9) L(x) = x(x^2 - 6x + 9) - 8(x^2 - 6x + 9) L(x) = x^3 - 6x^2 + 9x - 8x^2 + 48x - 72 L(x) = x^3 - 14x^2 + 57x - 72 The LCM of the two polynomials is (x-8)(x-3)^2 or x^3 - 14x^2 + 57x - 72.