Direct Ratio MCQ Quiz in मल्याळम - Objective Question with Answer for Direct Ratio - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Given:

\(\rm tan A = {3 \over 8},\)

Concept used:

TanA = SinA / CosA

Calculation:

\({3 \sin A + 2 \cos A} \over 3 \sin A - 2 \cos A\)

Dividing by CosA we get,

3 TanA + 2 / 3 TanA - 2

Now putting the value of Tan A we get,

⇒ \( (3 × {3\over8}+ 2)\over (3 × {3\over8}- 2)\)

⇒ \(- {25 \over7}\)

∴ The  correct option is 2

Given

a cot θ = b

Calculation

\(\frac{bcosθ - asinθ}{bcosθ + asinθ}\)

Dividing the whole equation by cosθ, we get:

⇒ (b - asinθ/cosθ)/(b + asinθ/cosθ)

⇒ (b - atanθ)/(b + atanθ)

⇒ (b - a × a/b)/(b + a × a/b) 

⇒ (b² - a²)/(b² + a²) 

The value of the expression is  (b2 - a2)/(b2 + a2) .

Given

tan x = 7/5

Formula used

tan x = P/B

sin x = P/H

cos x = B/H

Calculation

tan x = 7/5 = P/B

H² = P² + B²

H2 = 7² + 5²

H2 = 49 + 25 = 74

h = √74.

\(\frac{9 \sin x - \frac{42}{5}\cos x}{15 \sin x + 21 \cos x}\)

⇒ (9 × 7/√74 - 42/5 × 5/√74)/(15 × 7/√74 + 21 × 5/√74)

⇒ (21/√74)/(210/√74)

⇒ 21/210

⇒ 0.1

The value is 0.1

Given:

Cos A = 9/41 

Formula used:

 cot A = cos A/sin A

Cos² A + sin² A = 1

Calculation:

Cos2 A + sin2 A = 1

⇒ ( 9/41)² + sin2 A = 1

⇒  sin2 A = 1 - 81/1681

 ⇒ sin2 A = 1600/1681

⇒  sin A = 40/41

⇒ cot A = cos A/sin A

 ⇒ cot A = 9/40

∴ Option 1 is the correct answer.

Calculation:

cosec²θ - cot²θ = 1, 

cosec²θ = 1 + cot²θ 

So,

(secθ(1 + cot2θ)(cosec2θ - cot2θ)) / cosec3θ

⇒ (sec θ × cosec²θ × 1) / cosec³θ

⇒ (sec θ) / (cosec θ)

⇒ (1/cos θ) / (1/sinθ)

⇒ sinθ / cosθ

⇒ tan θ = 1/cot θ = 1/(4/3) = 3/4

∴ The corrct answer is option (1).

Given:

sin θ = \(1 \over 2\)

Concept used:

Calculation:

sin θ = \(1 \over 2\)

⇒ sinθ = sin 30°

So, θ = 30°

3cos θ - 4 cos3 θ

⇒ 3 × \(\sqrt3 \over 2\) - 4 × \(3\sqrt3 \over 8\)

⇒ \({3\sqrt3 \over 2} - \frac{3\sqrt3}{2}\)

⇒ 0

∴ The required answer is 0.

Given:

Tan α = 6

Concept used:

Tan θ = P/B

Sec θ = H/B

Where, P = perpendicular; B = base; H = hypotenuse

Formula used:

Pythagorean theorem:

H² = P² + B²

Where, P = perpendicular; B = base; H = hypotenuse

Calculation:

Tan α = 6/1 = P/B

Using Pythagorean theorem:

⇒ H2 = P2 + B2

⇒ H² = (6)² + (1)²

⇒ H = √37

Sec α = H/B = √37

∴ The correct answer is √37. 

Given:

4 tan θ = 3

Calculation:

Applying Pythagoras Theorem

AC² = AB² + BC²

AC² = 3² + 4²

AC = 5

sinθ = 3/5

cosθ =4/5

\(\frac{4 \sin θ-\cos θ+1}{4 \sin θ+\cos θ-1}\)

Put the above values

⇒ \(\frac{4 \times 3/5-4/5+1}{4 \times 3/5+4/5-1}\)

⇒ \(\frac{12/5-4/5+1}{12/5+4/5-1}\) = \(\frac{(12-4+5)/5}{(12 +4 - 5)/5}\)

⇒ \(\frac{13/5}{11/5}\) = \(\frac{13}{11}\) 

∴ Option 4 is the correct answer.

Given:

7tan θ = 3

Calculation:

\(\frac{5 \sin \theta - \cos \theta}{5 \sin \theta + 2 \cos \theta}\)

Divide by 'cos θ' in both numerator and denominator,

⇒ \(\frac{5 (\sin \theta/ \cos \theta) -1}{5 (\sin \theta/ \cos \theta) + 2 }\)

⇒ \(\frac{5\ tan \theta -1}{5 \ tan \theta + 2 }\)

⇒ [5 × 3/7 - 1] / [5 × 3/7 + 2]

⇒ [15/7 - 1] / [15/7 + 2]

⇒ [(15 - 7)/7] / [(15 + 14)/7]

⇒ [8/7] / [29/7]

⇒ 8/29

∴ The correct answer is option (1).