Let, the height of the post = AB = ‘h’ m From ΔACB, ∠ACB = 30° [∵ given] tan ∠ACB = perpendicular/base ⇒ tan 30° = AB/BC ⇒ 1/√3 = h/BC ⇒ BC = h√3 ... (1) From ΔADB, ∠ADB = 60° [∵ given] tan ∠ADB = perpendicular/base ⇒ tan 60° = AB/BD ⇒ √3 = h/BD ⇒ BD = h/√3 ... (2) According to the question: BC - BD = 5 ⇒ h√3 - h/√3 = 5 ⇒ 2h/√3 = 5 ⇒ h = (5√3)/2 m
I. x²= 961
II. y= √961
I: x² - 10x + 21 = 0
II: 4y² - 16y + 15 = 0
I. 96y² - 76y – 77 = 0
II. 6x² - 19x + 15 = 0
A and B are the roots of equation x2 - 13x + k = 0. If A - B = 5, what is the value of k?
I. 10x2 + 33x + 9 = 0
II. 2y2 + 13y + 21 = 0
Equation 1: x² - 250x + 15625 = 0
Equation 2: y² - 240y + 14400 = 0
I. 7x² + 27x + 18 = 0
II. 19y² - 27y + 8 = 0
I. 14p2 – 135p + 81 = 0
II. 7q2 – 65q + 18 = 0
Solve the quadratic equations and determine the relation between x and y:
Equation 1: x² - 34x + 288 = 0
Equation 2: y² - 29y + 210 = 0
I. 8a2 – 22a+ 15 = 0
II. 12b2 - 47b + 40=0
