Question

Please help with this qn asap!! tyy!!​

Answer 1

`\textsf{(i)} \quad \left(6-√(3) \right)\; \sf cm`

`\textsf{(ii)} \quad \left(37-12√(3)\right)\; \sf cm^2`

Explanation:

Part (i)

`\boxed{\begin{minipage}{7.6 cm}\underline{Area of a triangle} \\\\$A=(1)/(2)ab \sin C$\\\\where:\\ \phantom{ww}$\bullet$ $C$ is the angle. \\ \phantom{ww}$\bullet$ $a$ and $b$ are the sides enclosing the angle. \\\end{minipage}}`

Given:

• 
  `\textsf{Area}=(1)/(4)(12+9√(3))`

• 
  `a=(√(3)+2)`

• 
  `C=60^(\circ)`

Substitute the given values into the formula to find AC (side b).

`\begin{aligned}\textsf{Area}&=(1)/(2)ab \sin C\\\\\implies (1)/(4)(12+9√(3))&=(1)/(2)(√(3)+2)b \sin 60^(\circ)\\\\(1)/(4)(12+9√(3))&=(1)/(2)(√(3)+2) (√(3))/(2)b\\\\(1)/(4)(12+9√(3))&=(√(3))/(4)(√(3)+2)b\\\\12+9√(3)&=√(3)(√(3)+2)b\\\\12+9√(3)&=(3+2√(3))b\\\\b&=(12+9√(3))/(3+2√(3))\\\\b&=(12+9√(3))/(3+2√(3))\cdot (3-2√(3))/(3-2√(3))\end{aligned}`

`\begin{aligned}b&=(36-24√(3)+27√(3)-54)/(9-6√(3)+6√(3)-12)\\\\b&=(-18+3√(3))/(-3)\\\\b&=6-√(3)\end{aligned}`

Therefore, the length of AC is (6 - √3) cm.

Part (ii)

`\boxed{\begin{minipage}{6 cm}\underline{Cosine Rule} \\\\$c^2=a^2+b^2-2ab \cos C$\\\\where:\\ \phantom{ww}$\bullet$ $a, b$ and $c$ are the sides.\\ \phantom{ww}$\bullet$ $C$ is the angle opposite side $c$. \\\end{minipage}}`

Given:

• 
  `c = BC`
• 
  `a = AB=(√(3)+2)`
• 
  `b = AC=(6-√(3))`
• 
  `C = \angle BAC=60^(\circ)`

Substitute the given values into the formula to find BC²:

`\begin{aligned}c^2&=a^2+b^2-2ab \cos C\\\\\implies BC^2&=AB^2+AC^2-2(AB)(AC) \cos C\\\\BC^2&=(√(3)+2)^2+(6-√(3))^2-2(√(3)+2)(6-√(3)) \cos 60^(\circ)\\\\BC^2&=7+4√(3)+39-12√(3)-2(9+4√(3)) \cdot (1)/(2)\\\\BC^2&=7+4√(3)+39-12√(3)-9-4√(3) \\\\BC^2&=37-12√(3)\end{aligned}`

Therefore, BC² is (37 - 12√3) cm².

Answer 2

• i) AC = 6 - √3 cm
• ii) BC² = 37 - 12√3 cm²

----------------------

Given triangle with

• Area of 1/4(12 + 9√3),
• Side AB = √3 + 2,
• Angle BAC is 60°.

Part i

Find the side AC.

Use area formula:

• A = ab sin∠C / 2, where a, b are adjacent sides and C is angle between a and b.

Substitute and solve for AC:

• 1/4(12 + 9√3) = (√3 + 2) AC sin(60°) / 2
• 1/4(12 + 9√3) = AC (√3 + 2) (√3/2) / 2
• (12 + 9√3)/4 = AC √3(√3 + 2)/4
• 12 + 9√3 = AC√3(√3 + 2)
• AC = (12 + 9√3) /√3(√3 + 2)
• AC = (4√3 + 9) / (√3 + 2)
• AC = (4√3 + 9)(√3 - 2) / (√3 + 2)(√3 - 2)
• AC = (4√3√3 + 9√3 - 8√3 - 18) / (3 - 4)
• AC = (4*3 + √3 - 18) / ( - 1)
• AC = - (√3 - 6)
• AC = 6 - √3

Part ii

Use the law of cosines

• BC² = AB² + AC² - 2 × AB × AC × cos ∠A
• BC² = (√3 + 2)² + (6 - √3)² - 2(√3 + 2)(6 - √3) cos 60°
• BC² = 3 + 4 + 4√3 + 36 + 3 - 12√3 - 2(6√3 - 3 + 12 - 2√3) / 2
• BC² = 46 - 8√3 - (4√3 + 9)
• BC² = 46 - 8√3 - 4√3 - 9
• BC² = 37 - 12√3