मराठी

Find the angle θ, when 0° < θ < 90°. 3 tan2 (θ + 10°) = 1, 0° < θ < 90° - Mathematics

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प्रश्न

Find the angle θ, when 0° < θ < 90°.

3 tan2 (θ + 10°) = 1, 0° < θ < 90°

बेरीज
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उत्तर

Given:

3 tan2 (θ + 10°) = 1, 0° < θ < 90°

Step 1: Simplify

3 tan2 (θ + 10°) = 1

tan2 (θ + 10°) = `1/2`   ...[Divide both sides by 3]

Step 2: tan (θ + 10°) = ± `1/sqrt3`

Since 0° < θ < 90°

10° < θ + 10° < 100°

In the interval 0° to 90°, tangent is positive.

In the interval 90° to 100°, tangent is negative.

both + `1/sqrt3 and 1/sqrt3` 

tan (θ + 10°) = `1/sqrt3`

θ + 10° = 30°

θ = 20°    ...[0° < 20° < 90°]

tan(θ + 10°) = − `1/sqrt3` 

This happens at 150∘, 330∘,...

but in 10° < θ + 10° < 100°, there is no such value this case is not valid.

θ = 20°

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पाठ 19: Trigonometry - MISCELLANEOUS EXERCISE [पृष्ठ २३९]

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बी निर्मला शास्त्री Mathematics [English] Class 9 ICSE
पाठ 19 Trigonometry
MISCELLANEOUS EXERCISE | Q III. 1. | पृष्ठ २३९
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