Question

In the figure below, \(\Delta A C B\) is a right triangle with legs of length \(a\) units and \(b\) units, where \(0 < a < b,\) and hypotenuse of length \(c\) units. The triangles \(\triangle Y C A, \Delta Z B A,\) and \(\Delta X C B\) are equilateral. The area of an equilateral triangle with sides \(x\) units long is \(\frac{\sqrt{3}}{4} x^{2}\) square units. (See figure) If \(b=2 a,\) what is \(\tan (\angle A B C) ?\) F. 2 G. \(\frac{1}{2}\) H. \(\frac{1}{\sqrt{5}}\) J. \(\frac{2}{\sqrt{5}}\) K. \(\sqrt{5}\)

Short answer

Answer: The value of \(\tan(\angle ABC)\) given the mentioned conditions is J. \(\frac{2}{\sqrt{5}}\).

Step-by-step solution

Find the side length of equilateral triangles

Given that \(\triangle YCA\) and \(\triangle ZBA\) are equilateral triangles, their side lengths can be determined by analyzing the right triangle \(\triangle ACB\). Since \(YC = a\) and \(ZA = b\), also \(AY = AZ = YC = b\), and \(BZ = BA = ZA = a\). This means that the right triangle \(\triangle ACB\) has legs of length \(a+y\) and \(b+x\), hypotenuse of length \(c+z\) and angle \(ABC\).

Find \(c\) using the Pythagorean Theorem

Since \(\triangle ACB\) is a right triangle, we can apply the Pythagorean theorem to find the hypotenuse \(c\). From Step 1, we know that: \((a+y)^2 + (b+x)^2 = (c+z)^2\) We're also given that \(b = 2a\). Plugging that into the equation, we have: \((a+y)^2 + (2a + x)^2 = (c+z)^2\)

Calculate \(\tan(\angle ABC)\)

We know that \(\tan(\angle ABC) = \frac{AC}{BC} =\frac{a+y}{b+x}\). To find \(\tan(\angle ABC)\), we need to solve for \(y\) and \(x\). From Step 1, we know that both \(y\) and \(x\) are equal to the side lengths of the equilateral triangles. Since the side length of the equilateral triangles is given in terms of \(a\) and \(b\), we can express \(y\) and \(x\) in terms of \(a\). Using the area formula for an equilateral triangle, \(A=\frac{\sqrt{3}}{4}x^2\), we can find \(y\) and \(x\) as follows: \(y = \frac{\sqrt{3}}{4}(a^2)\) and \(x = \frac{\sqrt{3}}{4}(b^2)\) Substitute the values of \(y\) and \(x\) in terms of \(a\) and \(b\) into the equation for \(\tan(\angle ABC)\): \(\tan(\angle ABC) =\frac{a+\frac{\sqrt{3}}{4}(a^2)}{2a + \frac{\sqrt{3}}{4}(2a)^2}\) Simplify the equation for \(\tan(\angle ABC)\): \(\tan(\angle ABC) =\frac{1+\frac{\sqrt{3}}{4}a}{2 + \frac{\sqrt{3}}{4}(4a)}= \frac{1+\frac{\sqrt{3}}{4}a}{2 + \sqrt{3}a}\) Now multiplying the numerator and the denominator by \((2-\sqrt{3}a)\) we have: \(\tan(\angle ABC) = \frac{(1+\frac{\sqrt{3}}{4}a)(2 - \sqrt{3}a)}{(2 + \sqrt{3}a)(2 - \sqrt{3}a)}\) After simplifying the expression, we get: \(\tan(\angle ABC) =\frac{2-\sqrt{3}a}{2-\sqrt{3}a}\cdot\frac{1+\frac{\sqrt{3}}{4}a}{1- \frac{3}{4}a}\) \(\tan(\angle ABC) =\frac{1+\frac{\sqrt{3}}{4}a}{1- \frac{3}{4}a}\) Now we have \(\tan(\angle ABC)\) in terms of \(a\). We know that \(b=2a\), therefore \(\frac{b}{2}=a\). Finally substituting for \(a\) with \(\frac{b}{2}\). \(\tan(\angle ABC) =\frac{1+\frac{\sqrt{3}}{4}(\frac{b}{2})}{1- \frac{3}{4}(\frac{b}{2})}=\frac{1+ \frac{\sqrt{3}}{8}b}{1 - \frac{3}{8}b}\) Upon comparing the values of \(\tan(\angle ABC)\) with the given choices, we find the answer is: J. \(\frac{2}{\sqrt{5}}\)

Key concepts

Trigonometry

Trigonometry is a branch of mathematics that revolves around the study of triangles, particularly right-angled triangles, and the relationships between their sides and angles.
One of the basic trigonometric functions is the tangent, often abbreviated as 'tan.' It is defined as the ratio of the opposite side to the adjacent side of a particular angle in a right triangle. In mathematical terms, for a right triangle with an angle \theta, \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \).
In our exercise, we're asked to find \( \tan(\angle ABC) \), which requires us to determine the lengths of the sides opposite and adjacent to \( \angle ABC \).
The tangent function is crucial in various fields such as engineering, physics, and even in everyday situations like determining the height of a structure using a measured angle from a certain distance. Understanding how to apply the tangent function to solve for angles and sides of triangles is fundamental in ACT Math Prep and all trigonometry courses.

Pythagorean Theorem

The Pythagorean Theorem is a cornerstone of geometry, specifically applicable to right-angled triangles. It expresses a clear relationship between the three sides of a right triangle: the two legs (a and b) and the hypotenuse (c).
The theorem states that the sum of the squares of the legs' lengths is equal to the square of the length of the hypotenuse. In formulaic terms, \( a^2 + b^2 = c^2 \).
Using this theorem, we can solve for any missing side length of a right triangle as long as the other two lengths are known. This concept is vital for solving more complex problems in trigonometry and ACT Math Prep.
In our exercise, the Pythagorean Theorem is used in conjunction with the given information to deduce the relationship between the various side lengths, which is then used as a stepping stone to calculate the tangent of the given angle \( \angle ABC \).
A solid understanding of the Pythagorean Theorem is not only beneficial for standardized tests but also for practical applications in science, technology, and various fields of mathematics.

Equilateral Triangle Area

The area of an equilateral triangle, which is a triangle with all three sides of equal length, can be calculated using a specific formula derived from its geometry.
For an equilateral triangle with side length \( x \), the formula for its area \( A \) is \( A = \frac{\sqrt{3}}{4} x^2 \).
This formula is crucial in solving our exercise as it allows us to find the length of the sides of the attached equilateral triangles, which are needed to eventually solve for the tangent of \( \angle ABC \). The square in the formula indicates that the area is directly proportional to the square of the side length, highlighting the geometric properties of equilaterality and symmetry.
In ACT Math Prep, understanding how to derive and apply this formula is essential for solving a variety of problems involving areas of triangles. Moreover, the elegance and simplicity of the formula make it easy to remember and utilize in many geometric and algebraic contexts.