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secondary 2 | Maths
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hi i need help
Date Posted: 3 years ago
Views: 190
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Hope this helps
Date Posted:
3 years ago
△ABC is right-angled.
AB² = BC² + AC² (Pythagoras' Theorem)
10² = 6² + AC²
AC² = 10² - 6² = 100 - 36 = 64
AC = √64 = 8
Now AD = DC (shown in diagram)
Since AC = AD + DC, DC = ½ AC
DC = ½(8) = 4
Using trigo, cos(∠ABC) = adjacent/hypotenuse = BC/AB
cos(∠ABC) = 6/10 = 3/5
∠ABC = cos-¹(3/5)
Next, since △DBC is also right-angled,
Using trigo, tan(∠DBC) = opposite/adjacent = DC/BC
tan(∠DBC) = 4/6 = 2/3
∠DBC = tan-¹(⅔)
θ = ∠ABD = ∠ABC - ∠DBC
= cos-¹(3/5) - tan-¹(⅔)
≈ 53.13° - 33.69°
= 19.44°
= 19.4° (1d.p)
AB² = BC² + AC² (Pythagoras' Theorem)
10² = 6² + AC²
AC² = 10² - 6² = 100 - 36 = 64
AC = √64 = 8
Now AD = DC (shown in diagram)
Since AC = AD + DC, DC = ½ AC
DC = ½(8) = 4
Using trigo, cos(∠ABC) = adjacent/hypotenuse = BC/AB
cos(∠ABC) = 6/10 = 3/5
∠ABC = cos-¹(3/5)
Next, since △DBC is also right-angled,
Using trigo, tan(∠DBC) = opposite/adjacent = DC/BC
tan(∠DBC) = 4/6 = 2/3
∠DBC = tan-¹(⅔)
θ = ∠ABD = ∠ABC - ∠DBC
= cos-¹(3/5) - tan-¹(⅔)
≈ 53.13° - 33.69°
= 19.44°
= 19.4° (1d.p)
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