Question

Point D is on side BC of equilateral ▲ABC. From point D, perpendicular line segments with lengths 4 and 8 inches are drawn meeting sides AB and AC at points R and T. Find the number of inches in the height of ▲ABC. [Hint: Draw a line segment from A to D.] answer pls ILL MARK BRAINLIEST

Answer 1

Correct response:

• The height of equilateral triangle ΔABC is 12 inches

Methods used to calculate the height of ΔABC

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;

$\displaystyle \frac{4}{sin(\theta)} = \mathbf{\frac{8}{sin(60^{\circ} - \theta) } }$

Therefore;

4·sin(60° - θ) = 8·sin(θ)

• sin(A - B) = sin(A)·cos(B) - cos(A)·sin(B)

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

$\displaystyle cot(\theta) = \mathbf{ \frac{5}{\sqrt{3} } }$

Therefore;

$\displaystyle tan(\theta) = \mathbf{ \frac{\sqrt{3} }{5} }$

$\displaystyle \theta = arctan \left(\frac{\sqrt{3} }{5 } \right)$

Length of a side of the equilateral triangle, L, is therefore;

$\displaystyle L = \mathbf{ \frac{4 \cdot \sqrt{3}}{3} + \frac{4}{tan(\theta)} }$

Therefore;

$\displaystyle L = \frac{4 \cdot \sqrt{3} }{3} +\frac{4}{\frac{\sqrt{3} }{5} } = 8 \cdot \sqrt{3}$

Height, h, of the equilateral triangle ΔABC is, h = L × sin(60°)

Therefore;

$\displaystyle h = 8 \cdot \sqrt{3} \times \frac{\sqrt{3} }{2} = \mathbf{ 12}$

• The height of the equilateral triangle, h = 12 inches

Learn more about trigonometric ratios here:

brainly.com/question/16829589

Answer 2

Answer: 24/√3

Step-by-step explanation: