Correct response:
- The height of triangle ΔABC is 12 inches.
Method used for finding the height
The given parameters are;
ΔABC is an equilateral triangle;
DR = 4 inches
DT = 8 inches
Required:
The height of ΔABC
Solution:
Let θ represent the angle formed by the line from A to D towards the 4 inches side, we have;
- [tex]AD = \dfrac{4}{sin(\theta)} = \mathbf{\dfrac{8}{sin(60^{\circ} - \theta)} }[/tex]
4·(sin(60° - θ) = 8·sin(θ)
4·(sin60°·cos(θ) - cos(60°)·sin(θ)) = 8·sin(θ)
Dividing by sin(θ) and multiplying by 2 gives;
4·(√3·cot(θ) - 1) = 8 × 2 = 16
√3·cot(θ) - 1 = 16 ÷ 4 = 4
[tex]cot(\theta) = \dfrac{4 + 1}{\sqrt{3} } = \dfrac{5}{\sqrt{3} } [/tex]
[tex]tan(\theta) = \mathbf{ \dfrac{\sqrt{3} }{5} }[/tex]
[tex]tan(\theta) = \dfrac{DR}{AR} [/tex]
Therefore;
[tex]AR = \mathbf{\dfrac{DR}{tan(\theta)} }[/tex]
Which gives;
[tex]AR = \dfrac{4}{\dfrac{\sqrt{3} }{5} } = 4 \times \dfrac{5}{\sqrt{3} } = \mathbf{ \dfrac{20}{\sqrt{3} } }[/tex]
[tex]RB = \mathbf{\dfrac{4}{tan(60^{\circ})}} = \dfrac{4}{\sqrt{3} } [/tex]
AB = AR + RB by angle addition property
Height = AB × sin(∠B) = (AR + RB) × sin(∠B)
Therefore;
[tex]Height \ of \ \Delta ABC, \ h = \left(\dfrac{20}{\sqrt{3} } + \dfrac{4}{\sqrt{3} } \right) \times \dfrac{\sqrt{3} }{2} =\dfrac{24}{\sqrt{3} } \times \dfrac{\sqrt{3} }{2} = \mathbf{ 12}[/tex]
- The height of triangle ΔABC, h = 12 inches
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