Tan() -- calculate the exact value w/o a calculator

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Homework Help Overview

The problem involves a right triangle ABC with a right angle at C, where side AC measures √3 cm and side BC measures 1 cm. The goal is to determine the angle A (α) without using a calculator, specifically focusing on the tangent function.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the sides and angles of the triangle, with some attempting to apply the tangent function and the Pythagorean theorem. Questions arise regarding the definitions of sine and cosine, particularly in relation to the triangle's dimensions.

Discussion Status

The discussion has seen various interpretations of the triangle's properties, with some participants providing insights into the relationships between angles and side lengths. There is a recognition of the need to clarify definitions and relationships, particularly concerning sine and cosine.

Contextual Notes

Participants are exploring the implications of the triangle's dimensions and the definitions of trigonometric functions, with some noting potential misunderstandings in the original poster's calculations.

Rectifier
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The problem
I have the triangle ABC where ∠C is 90°. The side AC is ## \sqrt{3} \ cm ## and the side BC is 1 cm. What is ∠A (α)?

The attempt
$$tan(a) = \frac{1}{\sqrt{3}}$$

The problem is that I have to calculate the exact value of α without any calculator.

I tried to calculate the third side by applying the pythagoras theorem. AB turned out to be 2 cm.

I even tried to write it like that $$tan(a) = \frac{sin(a)}{cos(a)} = \frac{1}{\sqrt{3}} \\ sin(a)=1 \\ a=90 \\ cos(a) = \sqrt{3}\\ $$

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Imagine an equilateral triangle with three sides of length s. Each angle is, of course, 60 degrees (\pi/3 radians). Drop a perpendicular from one vertex to the middle of the opposite side. One can show (by showing that the two triangles this divides the equilateral triangle into are congruent) that this new line is perpendicular to the opposite sides and bisects the angle at the vertex.

That is, the new line divides the equilateral triangle into two right triangles with angles of 60 degrees (\pi/3 radians) and 30 degrees (\pi/6 radians). One leg is of length s/2 and the hypotenuse is s so the other leg, the altitude of the original equilateral triangle, is of length \sqrt{s^2- s^2/4}= \frac{\sqrt{3}}{2}s.

So, what angle has tangent equal to \frac{1}{\sqrt{3}}?
 
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Recall that cos and sin range between -1 and 1, never outside that. So cos(a) = sqrt(3) can't be right.

You have the triangle having sides 1, 2, and sqrt(3). So the cos isn't sqrt(3). You are forgetting the definition of cos. Also, sin isn't 1.

What is the definition of sin? For this triangle it is 1 divided by something, but what?
 
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It's an half of an equilateral triangle with all sides = 2 units . All angles = 60 degrees . Your angle is a half angle = 30 degrees .

Or use sin(a) = 1/2 on original triangle . This is a well known result for 30 degrees .
 
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DEvens said:
Recall that cos and sin range between -1 and 1, never outside that. So cos(a) = sqrt(3) can't be right.

You have the triangle having sides 1, 2, and sqrt(3). So the cos isn't sqrt(3). You are forgetting the definition of cos. Also, sin isn't 1.

What is the definition of sin? For this triangle it is 1 divided by something, but what?

sin(a)= is the ratio of the opposite side and the hypothenuse
 
Oh!
I get it now
sin(a)= is the ratio of the opposite side and the hypothenuse and the hypothenuse is 2!

This means that $$ sin(a)= \frac{1}{2} $$ and then a=30°

Thank you to all of you!
 
HallsofIvy said:
Imagine an equilateral triangle with three sides of length s. Each angle is, of course, 60 degrees (\pi/3 radians). Drop a perpendicular from one vertex to the middle of the opposite side. One can show (by showing that the two triangles this divides the equilateral triangle into are congruent) that this new line is perpendicular to the opposite sides and bisects the angle at the vertex.

That is, the new line divides the equilateral triangle into two right triangles with angles of 60 degrees (\pi/3 radians) and 30 degrees (\pi/6 radians). One leg is of length s/2 and the hypotenuse is s so the other leg, the altitude of the original equilateral triangle, is of length \sqrt{s^2- s^2/4}= \frac{\sqrt{3}}{2}s.

So, what angle has tangent equal to \frac{1}{\sqrt{3}}?

I am sorry that I ignored your comment. I didnt see it in this thread when I first got replies for some reason.
 

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