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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
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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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पाठ 6: Trigonometry - Problem Set 6 [पृष्ठ १३८]

APPEARS IN

बालभारती Mathematics 2 [English] Standard 10 Maharashtra State Board
पाठ 6 Trigonometry
Problem Set 6 | Q 1.4 | पृष्ठ १३८

संबंधित प्रश्‍न

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


If tanθ = 1 then, find the value of

`(sinθ + cosθ)/(secθ + cosecθ)`


Prove that:

cos2θ (1 + tan2θ)


Prove that: `1/"sec θ − tan θ" = "sec θ + tan θ"`


Choose the correct alternative answer for the following question.
sin \[\theta\] cosec \[\theta\]= ?


Choose the correct alternative answer for the following question.

1 + tan2 \[\theta\]  = ?


Prove the following.

secθ (1 – sinθ) (secθ + tanθ) = 1


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.

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

Prove the following:

sec6x – tan6x = 1 + 3sec2x × tan2x


Choose the correct alternative: 
sinθ × cosecθ =?


Show that: 

`sqrt((1-cos"A")/(1+cos"A"))=cos"ecA - cotA"`


ΔAMT∼ΔAHE, construct Δ AMT such that MA = 6.3 cm, ∠MAT=120°, AT = 4.9 cm and `"MA"/"HA"=7/5`, then construct ΔAHE.


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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