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Find Acute Angles a and B, If Sin (A + 2b) = Sqrt3/2 Cos(A + 4b) = 0, a Greater than B - CBSE Class 10 - Mathematics

ConceptTrigonometric Ratios of an Acute Angle of a Right-angled Triangle

Question

Find acute angles A & B, if sin (A + 2B) = sqrt3/2 cos(A + 4B) = 0, A > B

Solution

Sin (A + 2B) = Sin 60°

Cos (A + 4B) = cos 90°

A + 2B = 60° …(i)

A + 4B = 90° …(ii)

Subtracting (ii) from (i)

A + 4B = 90°

(−A – 2B = −60)/(2B = 30°)  ∴ B = 15°

A + 4B = 90°

4B = 4(15°) = 4B = 60°

∴ A + 60° = 90° ∴ A = 30°

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Solution Find Acute Angles a and B, If Sin (A + 2b) = Sqrt3/2 Cos(A + 4b) = 0, a Greater than B Concept: Trigonometric Ratios of an Acute Angle of a Right-angled Triangle.
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