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

Choose the Correct Alternative Answer for the Following Question. When We See at a Higher Level, from the Horizontal Line, Angle Formed is ........ - Geometry Mathematics 2

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प्रश्न

Choose the correct alternative answer for the following question.

When we see at a higher level, from the horizontal line, angle formed is ........
 

पर्याय

  • angle of elevation.

  • angle of depression.

  • 0

  • straight angle.

MCQ

उत्तर

When we see at a higher level, from the horizontal line, angle formed is angle of elevation.

Hence, the correct answer is angle of elevation.

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Application of Trigonometry
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 6: Trigonometry - Problem Set 6 [पृष्ठ १३८]

संबंधित प्रश्‍न

If \[\sin\theta = \frac{7}{25}\], find the values of cosθ and tan​θ.


If \[\cot\theta = \frac{40}{9}\], find the values of cosecθ and sinθ.


If 5 secθ – 12 cosecθ = 0, find the values of secθ, cosθ, and sinθ.


Prove that:

cos2θ (1 + tan2θ)


Prove that:

(secθ - cosθ)(cotθ + tanθ) = tanθ.secθ.


Prove that:

\[\sec\theta + \tan\theta = \frac{\cos\theta}{1 - \sin\theta}\]

Prove the following.
(secθ + tanθ) (1 – sinθ) = cosθ


Prove the following.
sec2θ + cosec2θ = sec2θ × cosec2θ 


Prove the following.
cot2θ – tan2θ = cosec2θ – sec2θ


Prove the following.

\[\frac{\tan\theta}{\sec\theta + 1} = \frac{\sec\theta - 1}{\tan\theta}\]


Prove the following.

\[\frac{\tan^3 \theta - 1}{\tan\theta - 1} = \sec^2 \theta + \tan\theta\]

Choose the correct alternative: 
sinθ × cosecθ =?


If sinθ = `8/17`, where θ is an acute angle, find the value of cos θ by using identities.


In ΔPQR, ∠P = 30°, ∠Q = 60°, ∠R = 90° and PQ = 12 cm, then find PR and QR.


Prove that: (sec θ – cos θ) (cot θ + tan θ) = tan θ.sec θ

Proof: L.H.S. = (sec θ – cos θ) (cot θ + tan θ)

= `(1/square - cos θ) (square/square + square/square)` ......`[∵ sec θ = 1/square, cot θ = square/square and tan θ = square/square]`

= `((1 - square)/square) ((square + square)/(square  square))`

= `square/square xx 1/(square  square)`  ......`[(∵ square + square = 1),(∴ square = 1 - square)]`

 = `square/(square  square)`

= tan θ.sec θ

= R.H.S.

∴ L.H.S. = R.H.S.

∴ (sec θ – cos θ) (cot θ + tan θ) = tan θ.sec θ


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