Correct response:
Given:
ΔABC is a equilateral triangle
Point D is on side BC
Point R is on side AB
Point T is on side AC
DR = 4 inches
DT = 8 inches
Required:
The height in inches of equilateral triangle ΔABC.
Solution:
By sine rule, we have;
[tex]\displaystyle \frac{4}{sin(\theta)} = \mathbf{\frac{8}{sin(60^{\circ} - \theta) } }[/tex]
Therefore;
4·sin(60° - θ) = 8·sin(θ)
Therefore;
4·(sin(60°)·cosθ - cos(60°)·sin(θ)) = 8·sin(θ)
Dividing by sin(θ) gives;
4·(sin(60°)·cot(θ) - cos(60°)) = 8
sin(60°)·cot(θ) - cos(60°) = 8 ÷ 4 = 2
Multiplying by 2 gives;
√3·cot(θ) - 1 = 4
√3·cot(θ) = 4 + 1 = 5
[tex]\displaystyle cot(\theta) = \mathbf{ \frac{5}{\sqrt{3} } }[/tex]
Therefore;
[tex]\displaystyle tan(\theta) = \mathbf{ \frac{\sqrt{3} }{5} }[/tex]
[tex]\displaystyle \theta = arctan \left(\frac{\sqrt{3} }{5 } \right)[/tex]
Length of a side of the equilateral triangle, L, is therefore;
[tex]\displaystyle L = \mathbf{ \frac{4 \cdot \sqrt{3}}{3} + \frac{4}{tan(\theta)} }[/tex]
Therefore;
[tex]\displaystyle L = \frac{4 \cdot \sqrt{3} }{3} +\frac{4}{\frac{\sqrt{3} }{5} } = 8 \cdot \sqrt{3} [/tex]
Height, h, of the equilateral triangle ΔABC is, h = L × sin(60°)
Therefore;
[tex]\displaystyle h = 8 \cdot \sqrt{3} \times \frac{\sqrt{3} }{2} = \mathbf{ 12}[/tex]
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