Identify the relevant right-angled triangle.

You're on a roll — 8.1 Step 1: Identify the relevant right-angled triangle. Since D is directly above A, DAB is a right-angled triangle with DAB = 90^. Step 2: Use trigonometry in DAB. We are given AD = p and DBA = y. We want to find AB. In DAB: ( DBA) = OppositeAdjacent = (AD)/(AB) y = (p)/(AB) Step 3: Solve for AB. AB = (p)/( y) The length of AB in terms of p and y is AB = (p)/( y). 8.2 Step 1: Use the given information about ABC. We are given that A, B, C lie in the same horizontal plane, and AB = AC. This means ABC is an isosceles triangle. We are also given ABC = x. Since AB=AC, then ACB = x. We are given 2AD = BC. Since AD = p, then BC = 2p. Step 2: Construct an altitude in ABC. Draw a perpendicular line from A to BC, let the intersection point be M. Since ABC is isosceles with AB=AC, the altitude AM bisects BC. Therefore, BM = MC = (1)/(2) BC = (1)/(2)(2p) = p. Step 3: Use trigonometry in AMB. AMB is a right-angled triangle with AMB = 90^. In AMB: ( ABM) = AdjacentHypotenuse = (BM)/(AB) x = (p)/(AB) Step 4: Equate expressions for AB and simplify. From 8.1, we found AB = (p)/( y). From Step 3, we found AB = (p)/( x). Equating these two expressions for AB: (p)/( y) = (p)/( x) Divide both sides by p (since p 0): (1)/( y) = (1)/( x) Therefore, x = y. 8.3 Step 1: Use the relationship derived in 8.2. We have x = y. Step 2: Substitute the given value of x. Given x = 60^. 60^ = y Step 3: Calculate the value of 60^. We know that 60^ = (1)/(2). (1)/(2) = y Step 4: Solve for y. y = ((1)/(2)) y ≈ 26.565^ Rounding to one decimal place: y ≈ 26.6^ The size of y is 26.6^. Send me the next one 📸