# How to Solve Trigonometry Problems with Tangent

In right triangle, the tangent of a non-right angle is the ratio of the opposite leg to the adjacent leg. In terms of sine and cosine, the tangent function is also defined by the equation tan(x) = sin(x)/cos(x). The tangent of 90 degrees is undefined.

The inverse of tan(x) can be written as either arctan(x) or tan^{-1}(x). For example, if θ an acute angle of a right triangle, A is the length of the adjacent side, and B is the length of the opposite side, then tan(θ) = B/A and arctan(B/A) = θ. Both the tangent and inverse tangent function can be used to solve a variety of math problems in geometry and trigonometry.

## Example 1: Scaled Triangle

**Problem:** A right triangle with a height of y and a length of 2x is scaled horizontally by a factor of 1/2, so that the new length is x. The angle at the base of the original unscaled triangle was θ; the angle at the base of the scaled triangle is now 3θ/2. What is the value of θ?

**Solution:** The first relation we have is tan(θ) = y/(2x). The second relation we can get out of the triangle is tan(3θ/2) = y/x. If we eliminate the variables y and x from these two expressions, we get an equation in only the unknown quantity θ:

tan(θ) = (1/2)tan(3θ/2)

There are many ways to solve this equation. We could use numerical or graphical methods to generate an approximate solution, or we could use a more complicated algebraic technique to obtain an exact solution. Let's try the algebraic method. The first step is to create a new angle variable φ equal to θ/2. This gives us the equation

tan(2φ) = (1/2)tan(3φ)

Now we can use the double angle and triple angle tangent formulas:

tan(2z) = 2tan(z)/(1 - tan(z)^2)

tan(3z) = (3tan(z) - tan(z)^3)/(1 - 3tan(z)^2)

Applying these trigonometric identities to the φ equation we get

2tan(φ)/(1 - tan(φ)^2) = (3tan(φ) - tan(φ)^3)/(2 - 6tan(φ)^2)

Clearing the denominators gives us

4tan(φ) - 12tan(φ)^3 = tan(φ)^5 - 4tan(φ)^3 + 3tan(φ)

tan(φ)^5 + 8tan(φ)^3 - tan(φ) = 0

This quintic polynomial equation in tan(φ) has five solutions, three real and two imaginary. They are

tan(φ) = 0

tan(φ) = sqrt(sqrt(17) - 4)

tan(φ) = - sqrt(sqrt(17) - 4)

tan(φ) = * i *sqrt(sqrt(17) - 4)

tan(φ) = -

*sqrt(sqrt(17) - 4)*

**i**The only one that makes sense in the context of this problem is the positive real solution, tan(φ) = sqrt(sqrt(17) - 4). This gives us φ = arctan[sqrt(sqrt(17) - 4)] and θ = 2arctan[sqrt(sqrt(17) - 4)] as the exact solution. The approximate solution is θ ≈ 38.67 degrees.

## Example 2: The Snowman's Shadow, a Trigonometry Word Problem

**Problem:** At a certain latitude, the elevation angle of the sun (labeled θ in the diagram above) is decreasing at a rate of 0.3 degrees per minute starting from noon. At noon, the elevation angle is 90 degrees, i.e., the sun is directly overhead. Meanwhile, the height of the snow man is decreasing at a rate of 0.25 cm per minute starting from noon. If the height of the snowman is 300 cm at noon, how long is its shadow at 3:30 PM?

**Solution:** There are two factors that govern the length of the snowman's shadow: the elevation angle of the sun and the height of the snowman. The decreasing sun angle works to length the shadow, while the melting of the snow man works to shorten it. Although this may seem like a related rates problem in differential calculus, it can be solved with nothing more than algebra and trig.

First, the height H of the snowman at time = t minutes from noon is given by the equation

H(t) = 300 - 0.25t

Next, the length L of the snowman's shadow at time = t minutes from noon is given by the equation

H(t)/L(t) = tan(θ)

But θ is itself a function of t, θ(t) = 90 - 0.3t degrees. Using this fact, we get

L(t) = H(t)/tan(θ(t))

= (300 - 0.25t) / tan(90 - 0.3t)

The time 3:30 PMis 210 minutes from noon, so we can plug t = 210 into the equation above to obtain the length of the shadow

L(210) = (300 - 0.25*210)/tan(90 - 0.3*210)

= 247.5/tan(20)

= 680 cm

## Example 3: Overlapping Right Triangles

**Problem:** Two right triangles of the same size and shape are arranged as in the diagram above so that their right angles coincide. If the lengths of the segments are as indicated, what is the angle θ?

**Solution:** We can solve for the angle θ by finding the values of the angles x and y shown in the solution diagram below.

Each of the right triangles has leg of length 7 and 15. The angle y satisfies the relation tan(y) = 7/15, thus, y = arctan(7/15). The angle (180 - x) satisfies the relation tan(180-x) = 15/7, thus, x = 180 - arctan(15/7). This is equivalent to x = 90 + arctan(7/15).

The three interior angles of any triangle must always add up to 180 degrees, which means θ + x + y = 180. Since we know the values of x and y, we get

θ = 180 - x - y

= 90 - 2*arctan(7/15)

≈ 39.97 degrees

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