Solve For A A A : Tan ⁡ 2 A − ( 1 + 3 ) Tan ⁡ A + 3 = 0 \tan ^2 A - (1+\sqrt{3}) \tan A + \sqrt{3} = 0 Tan 2 A − ( 1 + 3 ) Tan A + 3 = 0 $0 \leq A \leq 360^{\circ}$26. A Tower And Flagstaff On Its Top Subtend Angles Of 30 ∘ 30^{\circ} 3 0 ∘ And 15 ∘ 15^{\circ} 1 5 ∘ At The Base.

Mar 8, 2025 by ADMIN 284 views

**Solving Trigonometric Equations: A Comprehensive Guide**

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Introduction

Trigonometric equations are a crucial part of mathematics, and solving them requires a deep understanding of trigonometric functions and their properties. In this article, we will focus on solving a specific trigonometric equation involving the tangent function. We will also explore a real-world problem involving the angles subtended by a tower and a flagstaff at the base.

Solving the Trigonometric Equation

The given trigonometric equation is:

$\tan ^2 A - (1+\sqrt{3}) \tan A + \sqrt{3} = 0$

To solve this equation, we can use the quadratic formula, which states that for an equation of the form $ax^2 + bx + c = 0$ , the solutions are given by:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In this case, we have:

$a = 1$ $b = -(1+\sqrt{3})$ $c = \sqrt{3}$

Substituting these values into the quadratic formula, we get:

$\tan A = \frac{(1+\sqrt{3}) \pm \sqrt{(1+\sqrt{3})^2 - 4(1)(\sqrt{3})}}{2(1)}$

Simplifying the expression under the square root, we get:

$\tan A = \frac{(1+\sqrt{3}) \pm \sqrt{4 + 2\sqrt{3}}}{2}$

$\tan A = \frac{(1+\sqrt{3}) \pm (2 + \sqrt{3})}{2}$

This gives us two possible values for $\tan A$ :

$\tan A = \frac{(1+\sqrt{3}) + (2 + \sqrt{3})}{2} = \frac{3 + 2\sqrt{3}}{2}$

$\tan A = \frac{(1+\sqrt{3}) - (2 + \sqrt{3})}{2} = \frac{-1}{2}$

Finding the Angles

Now that we have the possible values for $\tan A$ , we can find the corresponding angles using the inverse tangent function.

For the first value, we have:

$\tan A = \frac{3 + 2\sqrt{3}}{2}$

Using a calculator or a trigonometric table, we find that:

$A \approx 75.97^{\circ}$

For the second value, we have:

$\tan A = \frac{-1}{2}$

Using a calculator or a trigonometric table, we find that:

$A \approx 173.19^{\circ}$

Real-World Problem: Tower and Flagstaff

A tower and a flagstaff on its top subtend angles of $30^{\circ}$ and $15^{\circ}$ at the base. Find the height of the tower and the flagstaff.

Solution

Let the height of the tower be $h_1$ and the height of the flagstaff be $h_2$ . We can use the tangent function to relate the angles and the heights:

$\tan 30^{\circ} = \frac{h_1}{d}$

$\tan 15^{\circ} = \frac{h_2}{d}$

where $d$ is the distance from the base to the tower.

We can solve these equations simultaneously to find the values of $h_1$ and $h_2$ .

Solving the System of Equations

We can rewrite the first equation as:

$h_1 = d \tan 30^{\circ}$

$h_1 = d \frac{\sqrt{3}}{3}$

Similarly, we can rewrite the second equation as:

$h_2 = d \tan 15^{\circ}$

$h_2 = d \frac{1}{\sqrt{2 + \sqrt{3}}}$

Finding the Values of $h_1$ and $h_2$

We can substitute the values of $h_1$ and $h_2$ into the equations and solve for $d$ .

$d = \frac{h_1}{\tan 30^{\circ}}$

$d = \frac{h_2}{\tan 15^{\circ}}$

Equating the two expressions for $d$ , we get:

$\frac{h_1}{\tan 30^{\circ}} = \frac{h_2}{\tan 15^{\circ}}$

Substituting the values of $h_1$ and $h_2$ , we get:

$\frac{d \frac{\sqrt{3}}{3}}{\frac{\sqrt{3}}{3}} = \frac{d \frac{1}{\sqrt{2 + \sqrt{3}}}}{\frac{1}{\sqrt{2 + \sqrt{3}}}}$

Simplifying the expression, we get:

$d = \frac{h_2}{\tan 15^{\circ}}$

Substituting the value of $h_2$ , we get:

$d = \frac{h_2}{\frac{1}{\sqrt{2 + \sqrt{3}}}}$

$d = h_2 \sqrt{2 + \sqrt{3}}$

Finding the Value of $h_2$

We can substitute the value of $d$ into one of the original equations to find the value of $h_2$ .

$h_2 = d \tan 15^{\circ}$

$h_2 = h_2 \sqrt{2 + \sqrt{3}} \frac{1}{\sqrt{2 + \sqrt{3}}}$

$h_2 = h_2$

This is a trivial equation, and we can see that $h_2$ is equal to itself.

Finding the Value of $h_1$

We can substitute the value of $d$ into one of the original equations to find the value of $h_1$ .

$h_1 = d \tan 30^{\circ}$

$h_1 = h_2 \sqrt{2 + \sqrt{3}} \frac{\sqrt{3}}{3}$

$h_1 = \frac{h_2 \sqrt{3}}{3} \sqrt{2 + \sqrt{3}}$