A tree breaks due to a storm and the top touches the ground 30 meters away from its base, making a 30-degree angle with the ground. What was the height of the tree before it broke?
The height of the tree before it broke =AB+AC trigonometry (tan) = opposite/adjacent): tan 30° = AB/30 Since tan 30° = 1/√3 1 /√3 =AB/30 AB=30/√3 =10√3m AB = 10√3 Cos 30° =BC/AC √3 /2 =30 /AC AC =60 /√3 AC =20√3. Now the height of the tree=AB+AC =10√3 +20√3 =30√3
I. p2 - 19p + 88 = 0
II. q2 - 48q + 576 = 0
I. 20x² - 93x + 108 = 0
II.72y² - 47y - 144 = 0
I. 2x2– 25x + 33 = 0
II. 3y2+ 40y + 48 = 0
I. 27x6- 152x3+ 125 = 0
II. 216y6- 91y3+ 8 = 0
I.√(3x-17)+ x=15
II. y+ 135/y=24
I. 8a2 – 22a+ 15 = 0
II. 12b2 - 47b + 40=0
In each of these questions, two equations (I) and (II) are given.You have to solve both the equations and give answer
I. x2 – ...
I. 27x6-152x3+125=0
II. 216y6 -91y3+8=0
I. x2 - 20x + 96 = 0
II. y2 - 23y + 22 = 0
I. y² - 7 y – 18 = 0
II. x² + 10 x + 16 = 0
