Question

CASE STUDY FOR CLASS 10 MATHS.........​

Answer 1

Answer:

Step-by-step explanation:

A) Height of Section A = AC

In right angled ΔACO

∠ACO = 90

∠AOC = 30

tan 30 = opp / adj

1/√3 = AC / OC

1/√3 = AC / 24

AC = 24/√3

AC = 24 × √3/√3 × √3  -------------Rationalising numerator and denominator

AC = 24√3/3

AC = 8√3

AC = 8 × 1.73

AC = 13.84 m

Therefore Height of Section A is 13.84m

(a) 13.84 is correct

B) Height of section B = AB = BC - AC

In right angled Δ BCO

∠BCO = 90

∠BOC = 45

∴ ∠ CBO  = 45 --------- remaining angle of triangle

∴ ∠BCO = ∠CBO

∴ BC = OC --------- sides opposite to equal angles are equal

∴ BC = 24m

Height of section B = BC - AC

= 24 - 13.84

= 10.16m

∴ Height of section B is 10.16m

∴ (d) 10.16 is correct

Hope it Helps You :)

Answer 2

Given :- from image :-

• CD = 24 m .
• ∠ADC = 30°
• ∠BDC = 45°

To Find :-

• (A) The height of the section A = AC = ?
• (B) The height of the section B = BC = ?

Solution :-

(A)

In right angled ∆ACD we have,

→ ∠ADC = 30°

→ CD = 24 m

So,

→ tan 30° = AC/CD

→ 1/√3 = AC/24

→ AC = (24/√3)

rationalizing the denominator in RHS,

→ AC = (24/√3) × (24/√3)

→ AC = (24√3)/(√3 × √3)

→ AC = 24√3/3

→ AC = 8√3

→ AC = 8 × 1.73

→ AC = 13.84 m (a) (Ans.)

Hence, The height of the section A is equal to 13.84 m .

(B)

In right angled ∆BCD we have,

→ ∠BDC = 30°

→ CD = 24 m

So,

→ tan 45° = BC/CD

→ 1 = BC/24

→ BC = 24 m

then,

→ BA = BC - AC

→ BA = 24 - 13.84

→ BA = 10.16 m (d) (Ans.)

Hence, The height of the section B is equal to 10.16 m .

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