मराठी
महाराष्ट्र राज्य शिक्षण मंडळएस.एस.सी (इंग्रजी माध्यम) इयत्ता १० वी

Sin ⁡θ = 1/2, then θ = ? A) 30° B) 45° C) 60° D) 90°

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प्रश्न

`sin θ = 1/2`, then θ = ?

पर्याय

  • 30°

  • 45°

  • 60°

  • 90°

MCQ
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उत्तर

30°

Explanation:

`sin θ = 1/2`

∴ θ = 30°   ...`[sin 30^circ = 1/2]`

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पाठ 6: Trigonometry - Exercise

संबंधित प्रश्‍न

Prove the following identities, where the angles involved are acute angles for which the expressions are defined:

`(sin theta-2sin^3theta)/(2cos^3theta -costheta) = tan theta`


If `x/a=y/b = z/c` show that `x^3/a^3 + y^3/b^3 + z^3/c^3 = (3xyz)/(abc)`.


Prove the following trigonometric identities.

`(sec A - tan A)/(sec A + tan A) = (cos^2 A)/(1 + sin A)^2`


If x = r cos A cos B, y = r cos A sin B and z = r sin A, show that : x2 + y2 + z2 = r2


`(tan A + tanB )/(cot A + cot B) = tan A tan B`


If (cosec θ – sin θ) = a3 and (sec θ – cos θ) = b3, prove that a2b2(a2 + b2) = 1.


If `cos B = 3/5 and (A + B) =- 90° ,`find the value of sin A.


Write the value of cos1° cos 2°........cos180° .


If \[\sin \theta = \frac{4}{5}\] what is the value of cotθ + cosecθ? 


Prove the following identity :

`(tanθ + secθ - 1)/(tanθ - secθ + 1) = (1 + sinθ)/(cosθ)`


Prove the following identity : 

`(1 + tan^2A) + (1 + 1/tan^2A) = 1/(sin^2A - sin^4A)`


Prove the following identity : 

`(1 + sinθ)/(cosecθ - cotθ) - (1 - sinθ)/(cosecθ + cotθ) = 2(1 + cotθ)`


For ΔABC , prove that : 

`tan ((B + C)/2) = cot "A/2`


Prove that: 2(sin6θ + cos6θ) - 3 ( sin4θ + cos4θ) + 1 = 0.


Prove that cot θ. tan (90° - θ) - sec (90° - θ). cosec θ + 1 = 0.


Prove that sin (90° - θ) cos (90° - θ) = tan θ. cos2θ.


Without using trigonometric table, prove that
`cos^2 26° + cos 64° sin 26° + (tan 36°)/(cot 54°) = 2`


Prove the following identities:
`1/(sin θ + cos θ) + 1/(sin θ - cos θ) = (2sin θ)/(1 - 2 cos^2 θ)`.


If 1 + sin2α = 3 sinα cosα, then values of cot α are ______.


Find the value of sin2θ  + cos2θ

Solution:

In Δ ABC, ∠ABC = 90°, ∠C = θ°

AB2 + BC2 = `square`   .....(Pythagoras theorem)

Divide both sides by AC2

`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`

∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`

But `"AB"/"AC" = square and "BC"/"AC" = square`

∴ `sin^2 theta  + cos^2 theta = square` 


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