📚 Questions 81-90

✅ Solution

Step 1: Define variables

Let father’s present age = x years
Son’s present age = y years

Given: x + y = 45 …(1)

Step 2: Five years ago

Five years ago:
Father’s age = (x – 5) years
Son’s age = (y – 5) years

Given: (x – 5)(y – 5) = 124 …(2)

Step 3: Solve equations

From (1): y = 45 – x
Substitute in (2):
(x – 5)(45 – x – 5) = 124
(x – 5)(40 – x) = 124
40x – x² – 200 + 5x = 124
-x² + 45x – 200 = 124
-x² + 45x – 324 = 0
x² – 45x + 324 = 0

Step 4: Solve quadratic equation

x² – 45x + 324 = 0
x² – 36x – 9x + 324 = 0
x(x – 36) – 9(x – 36) = 0
(x – 36)(x – 9) = 0

x = 36 or x = 9

If x = 36, then y = 45 – 36 = 9
If x = 9, then y = 45 – 9 = 36 (not possible as father is older)

Answer: Father’s age = 36 years, Son’s age = 9 years

✅ Solution

Step 1: Volume of cylinder

Cylinder diameter = 12 cm ⇒ radius R = 6 cm
Height H = 15 cm

Volume of cylinder = πR²H
= π × 6² × 15
= π × 36 × 15 = 540π cm³

Step 2: Volume of cone

Let cone radius = r cm
Height h = 12 cm

Volume of cone = \(\frac{1}{3}πr²h\)
= \(\frac{1}{3}πr² × 12 = 4πr²\) cm³

Step 3: Equate volumes

Volume of cylinder = Volume of cone
540π = 4πr²
4r² = 540
r² = 135
r = √135 = √(9 × 15) = 3√15 cm

Diameter = 2r = 6√15 cm ≈ 6 × 3.873 = 23.238 cm

Answer: Diameter of cone = 6√15 cm ≈ 23.24 cm

✅ Solution

Step 1: Use area formula

Area of triangle with vertices (x₁,y₁), (x₂,y₂), (x₃,y₃):
Area = \(\frac{1}{2}|x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)|\)

Given: A(2,3), B(-1,0), C(2,-4)
x₁=2, y₁=3; x₂=-1, y₂=0; x₃=2, y₃=-4

Step 2: Substitute values

Area = \(\frac{1}{2}|2(0 – (-4)) + (-1)(-4 – 3) + 2(3 – 0)|\)
= \(\frac{1}{2}|2(4) + (-1)(-7) + 2(3)|\)
= \(\frac{1}{2}|8 + 7 + 6|\)
= \(\frac{1}{2}|21|\)
= 10.5 square units

Answer: Area = 10.5 square units

✅ Proof

Step 1: Start with LHS

LHS = \(\frac{\tan θ}{1 – \cot θ} + \frac{\cot θ}{1 – \tan θ}\)

Convert to sin and cos:
tan θ = sin θ/cos θ, cot θ = cos θ/sin θ

Step 2: Simplify first term

First term = \(\frac{\frac{\sin θ}{\cos θ}}{1 – \frac{\cos θ}{\sin θ}}\)
= \(\frac{\frac{\sin θ}{\cos θ}}{\frac{\sin θ – \cos θ}{\sin θ}}\)
= \(\frac{\sin θ}{\cos θ} × \frac{\sin θ}{\sin θ – \cos θ}\)
= \(\frac{\sin² θ}{\cos θ(\sin θ – \cos θ)}\)

Step 3: Simplify second term

Second term = \(\frac{\frac{\cos θ}{\sin θ}}{1 – \frac{\sin θ}{\cos θ}}\)
= \(\frac{\frac{\cos θ}{\sin θ}}{\frac{\cos θ – \sin θ}{\cos θ}}\)
= \(\frac{\cos θ}{\sin θ} × \frac{\cos θ}{\cos θ – \sin θ}\)
= \(\frac{\cos² θ}{\sin θ(\cos θ – \sin θ)}\)
= \(-\frac{\cos² θ}{\sin θ(\sin θ – \cos θ)}\)

Step 4: Add both terms

LHS = \(\frac{\sin² θ}{\cos θ(\sin θ – \cos θ)} – \frac{\cos² θ}{\sin θ(\sin θ – \cos θ)}\)
= \(\frac{1}{\sin θ – \cos θ}(\frac{\sin² θ}{\cos θ} – \frac{\cos² θ}{\sin θ})\)
= \(\frac{1}{\sin θ – \cos θ}(\frac{\sin³ θ – \cos³ θ}{\sin θ \cos θ})\)

Using a³ – b³ = (a-b)(a²+ab+b²):
= \(\frac{1}{\sin θ – \cos θ} × \frac{(\sin θ – \cos θ)(\sin² θ + \sin θ \cos θ + \cos² θ)}{\sin θ \cos θ}\)
= \(\frac{\sin² θ + \cos² θ + \sin θ \cos θ}{\sin θ \cos θ}\)
= \(\frac{1 + \sin θ \cos θ}{\sin θ \cos θ}\)
= \(\frac{1}{\sin θ \cos θ} + 1\)
= \(\sec θ \csc θ + 1\)

= RHS (Proved)