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Question
If the angles of a triangle are 30°, 60°, and 90°, then shown that the side opposite to 30° is half of the hypotenuse, and the side opposite to 60° is `sqrt(3)/2` times of the hypotenuse.
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Solution
Given : In ΔCAB, m∠A=90°, m∠B = 60°, M∠C=30°
To prove : i AB = `1/2`BC ii. AC = `sqrt(3)/2 BC`
Construction: Take a point 'D' on ray BA such that AB = AD. join point C to point D.
Proof: In ΔCBD,
AD= AB ....[By construction]
∴ A is the midpoint of seg BD ....(i)
Also, m∠CAB = 90° ....[Given]
∴ seg CA ⊥ seg BD .....(ii)
∴ seg CA is the perpendicular bisector of seg BD ....[From(i) and (ii)]
∴ CD = CB ...........[By perpendicular bisector theorem]
∴ ΔCDB is an isosceles triangle
∴ ∠CDB ≅ ∠CBD .....(iii)[By isosceles triangle theorem]
But,∠CBD = 60° ....(iv) [Given]
∴ ∠CDB = 60° ....[from (iii) and (iv)]
∴ ∠BCD = 60° .....[Remaining angle of a triangle ]
∴ ΔCDB is an equilateral triangle ....[All angle are 60°]
∴ BD = BC = CD ....(vi)[Sides of equilateral triabgle ]
AB = `1/2` BD .....(vi) [By construction]
AB = `1/2` BC . ...(vii) [ From (v) and (vi)]
In ΔCAB,
∠CAB = 90° ....[Given]
∴ BC2 = AC2+AB2 ............[ By pythagoras theorem]
∴` BC^2 = AC^2 + (1/2 BC)^2` ...[From (vii)]
∴`BC^2 = AC^2 +1/4 BC^2`
∴ `AC^2 = BC^2 -1/4 BC^2`
∴ `Ac^2 = (4BC^2-BC^32)/4`
∴ `AC^2 = (3BC^2)/4`
∴ `AC = sqrt(3)/2 BC` ...[ Taking square root on both sides]
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