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Answer

Find out the altitude of the equilateral triangle .

To proof

By using the trignometric identity.

\(tan\theta = \frac{Perpendicular}{base}\)

As shown in the diagram

and putting the values of the angles , base and perpendicular

\(tan 60^{\circ} = \frac{a}{4\sqrt{3}}\)

\(tan 60^{\circ} = \sqrt{3}\)

solving

\(\sqrt{3} = \frac{a}{4\sqrt{3}}\)

\(a = \sqrt{3}\times 4 \sqrt{3}\)

As

\(\sqrt{3}\times \sqrt{3} = 3\)

put in the above

a = 4 × 3

a = 12 units

The  length of the altitude of the equilateral triangle is 12 units .

Option (F) is correct .

Hence proved

Answer:

F. 12

Step-by-step explanation:

We have been given an image of a triangle and we are asked to find the length of the altitude of our given triangle.

Since we know that altitude of an equilateral triangle splits it into two 30-60-90 triangle.

We will use Pythagoras theorem to solve for the altitude of our given triangle.

\(\text{Leg}^2+\text{Leg}^2=\text{Hypotenuse}^2\)

Upon substituting our given values in above formula we will get,

\((4\sqrt{3})^2+a^2=(8\sqrt{3})^2\)

\(16*3+a^2=64*3\)

\(48+a^2=192\)

\(48-48+a^2=192-48\)

\(a^2=144\)

Upon taking square root of both sides we will get,

\(a=\sqrt{144}\)

\(a=12\)

Therefore, the length of the altitude of our given equilateral triangle is 12 units and option F is the correct choice.

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