ABC is a right angles triangle with base BC and height AB. The hypotenuse AC is four times the length of the perpendicular drawn to it from opposite vertex. What is tan C equal to?
2 – √3
Solving the Right Triangle Problem for Tan C
The question asks us to find the value of tan C in a right-angled triangle ABC, where the right angle is at B. We are given that BC is the base and AB is the height. A key piece of information is that the hypotenuse AC is four times the length of the perpendicular drawn from the opposite vertex (B) to the hypotenuse AC.
Setting up the Triangle and Given Information
Let's denote the lengths of the sides and the altitude:
- AB = h (height)
- BC = b (base)
- AC = c (hypotenuse)
Since it's a right triangle at B, by the Pythagorean theorem, \(h^2 + b^2 = c^2\).
Let BD be the perpendicular from vertex B to the hypotenuse AC. Let the length of this perpendicular be p. So, BD = p.
We are given the condition that the hypotenuse AC is four times the perpendicular BD: \(c = 4p\).
Using Area of the Triangle
The area of the right-angled triangle ABC can be calculated in two ways:
- Using base BC and height AB: Area \( = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times b \times h \).
- Using base AC and height BD (the perpendicular to the hypotenuse): Area \( = \frac{1}{2} \times \text{hypotenuse} \times \text{perpendicular} = \frac{1}{2} \times c \times p \).
Equating the two expressions for the area:
\[ \frac{1}{2}bh = \frac{1}{2}cp \] \[ bh = cp \]Now, substitute the given condition \(c = 4p\) into this equation:
\[ bh = (4p)p \] \[ bh = 4p^2 \]Relating Sides to Angles using Trigonometry
In the right-angled triangle ABC:
- \(\sin C = \frac{\text{opposite side}}{\text{hypotenuse}} = \frac{AB}{AC} = \frac{h}{c}\). So, \(h = c \sin C\).
- \(\cos C = \frac{\text{adjacent side}}{\text{hypotenuse}} = \frac{BC}{AC} = \frac{b}{c}\). So, \(b = c \cos C\).
Substitute these expressions for h and b into the equation \(bh = 4p^2\):
\[ (c \sin C)(c \cos C) = 4p^2 \] \[ c^2 \sin C \cos C = 4p^2 \]We know \(p = c/4\). Substitute this into the equation:
\[ c^2 \sin C \cos C = 4 \left(\frac{c}{4}\right)^2 \] \[ c^2 \sin C \cos C = 4 \left(\frac{c^2}{16}\right) \] \[ c^2 \sin C \cos C = \frac{c^2}{4} \]Since \(c\) is the length of the hypotenuse, \(c \ne 0\). We can divide both sides by \(c^2\):
\[ \sin C \cos C = \frac{1}{4} \]Using Trigonometric Identities
Recall the double angle identity for sine: \(\sin(2C) = 2 \sin C \cos C\).
From our equation, we have \(\sin C \cos C = \frac{1}{4}\). Multiply by 2:
\[ 2 \sin C \cos C = 2 \times \frac{1}{4} \] \[ \sin(2C) = \frac{1}{2} \]Finding the Angle C
Since C is an acute angle in a right triangle (\(0^\circ < C < 90^\circ\)), the angle 2C must be between \(0^\circ\) and \(180^\circ\). The angles whose sine is \(\frac{1}{2}\) in this range are \(30^\circ\) and \(150^\circ\).
So, \(2C = 30^\circ\) or \(2C = 150^\circ\).
This gives us two possibilities for angle C:
- \(C = \frac{30^\circ}{2} = 15^\circ\)
- \(C = \frac{150^\circ}{2} = 75^\circ\)
Determining the Value of Tan C
We need to find \(\tan C\). We have two possible values for C:
- If \(C = 15^\circ\), \(\tan C = \tan 15^\circ\).
- If \(C = 75^\circ\), \(\tan C = \tan 75^\circ\).
Let's calculate these values:
For \(\tan 15^\circ\): Using the identity \(\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}\) with \(A=45^\circ\) and \(B=30^\circ\):
\[ \tan 15^\circ = \tan(45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ} \] \[ = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \times \frac{1}{\sqrt{3}}} = \frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}} = \frac{\sqrt{3}-1}{\sqrt{3}+1} \]Rationalize the denominator:
\[ \frac{\sqrt{3}-1}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1} = \frac{(\sqrt{3}-1)^2}{(\sqrt{3})^2 - 1^2} = \frac{3 - 2\sqrt{3} + 1}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3} \]So, if \(C = 15^\circ\), \(\tan C = 2 - \sqrt{3}\).
For \(\tan 75^\circ\): Using the identity \(\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\) with \(A=45^\circ\) and \(B=30^\circ\):
\[ \tan 75^\circ = \tan(45^\circ + 30^\circ) = \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ} \] \[ = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \times \frac{1}{\sqrt{3}}} = \frac{\frac{\sqrt{3}+1}{\sqrt{3}}}{\frac{\sqrt{3}-1}{\sqrt{3}}} = \frac{\sqrt{3}+1}{\sqrt{3}-1} \]Rationalize the denominator:
\[ \frac{\sqrt{3}+1}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1} = \frac{(\sqrt{3}+1)^2}{(\sqrt{3})^2 - 1^2} = \frac{3 + 2\sqrt{3} + 1}{3 - 1} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3} \]So, if \(C = 75^\circ\), \(\tan C = 2 + \sqrt{3}\).
Considering the Interpretation of "Base" and "Height"
The problem states "with base BC and height AB". While in a right triangle at B, BC and AB are simply the legs, sometimes specifying BC as the 'base' might imply it is the longer leg, making the angle opposite to it (angle A) larger, and consequently angle C smaller than angle A. If BC > AB, then \(\angle A > \angle C\). Since \(A + C = 90^\circ\), \(A > C\) implies \(C < 45^\circ\). If we follow this interpretation, then \(C = 15^\circ\) is the appropriate angle.
If \(C = 15^\circ\), then \(\tan C = 2 - \sqrt{3}\).
Final Answer
Based on our derivation, the possible values for \(\tan C\) are \(2 - \sqrt{3}\) and \(2 + \sqrt{3}\). Considering the likely intended meaning where the base BC is longer than the height AB, leading to \(C < 45^\circ\), the value of \(\tan C\) is \(2 - \sqrt{3}\).
Revision Table: Right Triangle Trigonometry
| Concept | Formula/Relationship | Application in this Problem |
|---|---|---|
| Area of Triangle | \( \frac{1}{2} \times \text{base} \times \text{height} \) | \( \frac{1}{2}bh = \frac{1}{2}cp \) |
| Sine in Right Triangle | \( \sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} \) | \( \sin C = h/c \), \( \sin A = b/c \) |
| Cosine in Right Triangle | \( \cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} \) | \( \cos C = b/c \), \( \cos A = h/c \) |
| Tangent in Right Triangle | \( \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} \) | \( \tan C = h/b \) |
| Complementary Angles | If \(A+C=90^\circ\), then \( \sin A = \cos C \) and \( \cos A = \sin C \) | Used to relate \(\sin A\) and \(\cos C\). |
| Double Angle Identity | \( \sin(2\theta) = 2 \sin \theta \cos \theta \) | Used to simplify \( \sin C \cos C = 1/4 \) to \( \sin(2C) = 1/2 \) |
| Special Angle Tangent | \( \tan 15^\circ = 2 - \sqrt{3} \) \( \tan 75^\circ = 2 + \sqrt{3} \) |
Possible values for tan C. |
Additional Information: Perpendicular to Hypotenuse
In a right-angled triangle ABC, with the right angle at B, the altitude BD to the hypotenuse AC creates two smaller triangles, ABD and BCD. These smaller triangles are similar to the original triangle ABC and also similar to each other. This similarity provides useful geometric mean relationships:
- \(h^2 = AD \times c\) (where AB=h, AC=c, AD is projection of AB on AC)
- \(b^2 = CD \times c\) (where BC=b, AC=c, CD is projection of BC on AC)
- \(p^2 = AD \times CD\) (where BD=p)
These relationships can also be used to solve problems involving the altitude to the hypotenuse. In our specific problem, we used the area relationship, which is often simpler when the ratio of the hypotenuse to the altitude is given.
The condition \(c = 4p\) is specific and leads to the angle \(C\) being either \(15^\circ\) or \(75^\circ\). Other ratios would lead to different angles.
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