3a) Proof the following trigonometric theorems.

Okay Benoni, let's work through these trigonometry problems. 3a) Proof the following trigonometric theorems. i) = Step 1: Start with the right-hand side (RHS) of the equation. RHS = Step 2: Recall the identity = ( )/( ). Substitute this into the RHS. RHS = (( )/( )) Step 3: Cancel out from the numerator and denominator. RHS = Step 4: Since RHS = and LHS = , the identity is proven. LHS = RHS = ii) ^2 = 1 + ^2 Step 1: Start with the fundamental Pythagorean identity. ^2 + ^2 = 1 Step 2: Divide every term in the equation by ^2 . (^2 )/(^2 ) + (^2 )/(^2 ) = (1)/(^2 ) Step 3: Use the identities = ( )/( ) and = (1)/( ). (( )/( ))^2 + 1 = ((1)/( ))^2 ^2 + 1 = ^2 Step 4: Rearrange the terms to match the given equation. ^2 = 1 + ^2 3b) State three (3) conditions each for using: i) Sine Rule The Sine Rule ((a)/( A) = (b)/( B) = (c)/( C)) is used to solve triangles when you know: • Two angles and any one side (AAS or ASA). • Two sides and a non-included angle (SSA, which can sometimes lead to an ambiguous case with two possible triangles). • To find the area of a triangle when two sides and the included angle are known (Area = (1)/(2)ab C). ii) Cosine Rule The Cosine Rule (a^2 = b^2 + c^2 - 2bc A) is used to solve triangles when you know: • Two sides and the included angle (SAS). • All three sides (SSS). • To find the third side of a triangle when two sides and the included angle are known. 3c) Write down the values of the following using the triangle below. Let the top vertex be A, the left base vertex be B, and the right base vertex be C. Let D be the point on BC where the altitude from A meets the base. From the image, we have: Altitude AD = 22 Base segment BD = 11 Base segment DC = 14 Angle at C is x. Angle at B is y. First, we need to find the lengths of the hypotenuses in the two right-angled triangles formed by the altitude. For ADC: Using the Pythagorean theorem, AC^2 = AD^2 + DC^2. AC^2 = 22^2 + 14^2 AC^2 = 484 + 196 AC^2 = 680 AC = sqrt(680) = sqrt(4 × 170) = 2sqrt(170) For ADB: Using the Pythagorean theorem, AB^2 = AD^2 + BD^2. AB^2 = 22^2 + 11^2 AB^2 = 484 + 121 AB^2 = 605 AB = sqrt(605) = sqrt(121 × 5) = 11sqrt(5) Now we can find the required trigonometric values: i) x In right-angled ADC: x = AdjacentHypotenuse = (DC)/(AC) x = (14)/(2sqrt(170)) = (7)/(sqrt(170)) To rationalize the denominator, multiply by sqrt(170)sqrt(170): x = 7sqrt(170)170 x = 7sqrt(170)170 ii) y In right-angled ADB: y = OppositeHypotenuse = (AD)/(AB) y = (22)/(11sqrt(5)) = (2)/(sqrt(5)) To rationalize the denominator, multiply by sqrt(5)sqrt(5): y = 2sqrt(5)5 y = 2sqrt(5)5 iii) x In right-angled ADC: x = OppositeAdjacent = (AD)/(DC) x = (22)/(14) x = (11)/(7) x = (11)/(7) 3 done, 2 left today. You're making progress.