J's answer to Ivey's Secondary 4 A Maths Singapore question.
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Visualisation. Explanation in comments.
Date Posted:
3 years ago
Triangle BON, DOM, ADB, ACB and CDB are similar triangles.
∠ACB = θ
For triangle BON ,
sin θ = opposite/hypotenuse = OB/BN = BN / 6
BN = 6 sin θ
For triangle ACB,
cos θ = adjacent / hypotenuse = AC / AB = AC / 6
AC = 2 cos θ
AM
= CM + AC
= BN + AC (since CBMN is a rectangle, CM = BN)
= 6 sin θ + 2 cos θ
But now, look at triangle AOM.
sin ∠AOM = opposite/hypotenuse = AM / OA
sin (θ + α) = (6 sin θ + 2 cos θ) / OA
OA sin (θ + α) = 6 sin θ + 2 cos θ
How to find OA? Look at triangle AOB.
OA² = OB² + AB² (Pythagoras's Theorem)
OA² = 6² + 2²
OA = √(6² + 2²) = √40
How to find α ?
tan α = opposite / adjacent = AB/OB = 2/6 = ⅓
α = tan-¹ (⅓)
(The basic angle)
So, putting these together,
√(6² + 2²) sin (θ + tan-¹(⅓) ) = 6 sin θ + 2 cos θ
√40 sin (θ + tan-¹(⅓)) = 6 sin θ + 2 cos θ
∠ACB = θ
For triangle BON ,
sin θ = opposite/hypotenuse = OB/BN = BN / 6
BN = 6 sin θ
For triangle ACB,
cos θ = adjacent / hypotenuse = AC / AB = AC / 6
AC = 2 cos θ
AM
= CM + AC
= BN + AC (since CBMN is a rectangle, CM = BN)
= 6 sin θ + 2 cos θ
But now, look at triangle AOM.
sin ∠AOM = opposite/hypotenuse = AM / OA
sin (θ + α) = (6 sin θ + 2 cos θ) / OA
OA sin (θ + α) = 6 sin θ + 2 cos θ
How to find OA? Look at triangle AOB.
OA² = OB² + AB² (Pythagoras's Theorem)
OA² = 6² + 2²
OA = √(6² + 2²) = √40
How to find α ?
tan α = opposite / adjacent = AB/OB = 2/6 = ⅓
α = tan-¹ (⅓)
(The basic angle)
So, putting these together,
√(6² + 2²) sin (θ + tan-¹(⅓) ) = 6 sin θ + 2 cos θ
√40 sin (θ + tan-¹(⅓)) = 6 sin θ + 2 cos θ
Sorry, ∠CAB = θ, not ∠ACB
Thx for clearing my doubts!! I got it now
No prob
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