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Tamil Nadu Board of Secondary EducationSSLC (English Medium) Class 10
Textbook Solutions8443
Concept Notes & Videos328
Syllabus

Angles of Elevation and Depression

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Topics

  • Definition:  Line of Sight
  • Definition: Angles of Elevation
  • Definition: Angle of Depression
CBSE: Class 10
Maharashtra State Board: Class 10
CISCE: Class 10

Definition: Line of Sight

The straight line joining the eye of the observer to the point on the object being viewed.

CBSE: Class 10
Maharashtra State Board: Class 10
CISCE: Class 10

Definition: Angle of Elevation

The angle between the line of sight and the horizontal through the observer’s eye, when the object is above the level of the observer’s eye.

CBSE: Class 10
Maharashtra State Board: Class 10
CISCE: Class 10

Definition: Angle of Depression

The angle between the line of sight and the horizontal through the observer’s eye, when the object is below the level of the observer’s eye.

Video Tutorials

We have provided more than 1 series of video tutorials for some topics to help you get a better understanding of the topic.

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Shaalaa.com | Application of Trigonometry part 1 (Concepts)

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Shaalaa.com


Application of Trigonometry part 1 (Concepts) [00:13:33]
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Application of Trigonometry part 1 (Concepts)
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Application of Trigonometry part 2 (Examples)
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Application of Trigonometry part 3 (Examples)
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Application of Trigonometry part 4 (Examples)
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Application of Trigonometry part 5 (Examples)
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Application of Trigonometry part 6 (Examples)
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Application of Trigonometry part 7 (Examples)
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Related QuestionsVIEW ALL [27]

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

Prove the following.
sec2θ + cosec2θ = sec2θ × cosec2θ 

Prove the following.

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

Prove that:

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

Prove that:

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

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

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

Prove that:

\[\sec^4 A\left( 1 - \sin^4 A \right) - 2 \tan^2 A = 1\]
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