Answer:
Height = 12 in
Step-by-step explanation:
From inspection of the diagram of the given information:
- ΔCTD and ΔBRD are both right angles with given bases TD and RD.
- ΔABC is an equilateral triangle ⇒ ∠TCD = ∠RBD = 60°
- Side length of ΔABC is equal to the sum of the hypotenuses of ΔCTD and ΔBRD.
Use the sine trigonometric ratio to calculate the hypotenuse of ΔCTD (CD):
[tex]\implies \sf \sin(\theta)=\dfrac{\textsf{side opposite the angle}}{hypotenuse}[/tex]
[tex]\implies \sf \sin 60^{\circ}=\dfrac{8}{CD}[/tex]
[tex]\implies \sf CD=\dfrac{16 \sqrt{3}}{3}[/tex]
As ΔCTD and ΔBRD are similar triangles, and RD is half TD, then BD is half CD:
[tex]\implies \sf BD=\dfrac{1}{2}CD[/tex]
[tex]\implies \sf BD=\dfrac{1}{2} \cdot \dfrac{16 \sqrt{3}}{3}[/tex]
[tex]\implies \sf BD=\dfrac{8 \sqrt{3}}{3}[/tex]
Therefore, the side length of the equilateral triangle ABC is:
[tex]\implies \sf CB=CD+BD[/tex]
[tex]\implies \sf CB=\dfrac{16 \sqrt{3}}{3}+\dfrac{8 \sqrt{3}}{3}[/tex]
[tex]\implies \sf CB=8 \sqrt{3}[/tex]
Height of an equilateral triangle formula
[tex]\sf h=\dfrac{\sqrt{3}}{2}s \quad \textsf{(where s is the side length)}[/tex]
Substitute the found side length CB into the formula and solve for h:
[tex]\implies \sf h=\dfrac{\sqrt{3}}{2} \cdot 8\sqrt{3}[/tex]
[tex]\implies \sf h=\dfrac{8\sqrt{3}\sqrt{3}}{2}[/tex]
[tex]\implies \sf h=\dfrac{8(3)}{2}[/tex]
[tex]\implies \sf h=\dfrac{24}{2}[/tex]
[tex]\implies \sf h=12\:\:in[/tex]
