How Can You Solve a Triangle Using the Law of Sines Without a Calculator?

AI Thread Summary
To solve Triangle ABC using the Law of Sines without a calculator, the angles can be determined from the tangent values given. Since tan B equals 1, angle B is 45 degrees. For angle A, with tan A equal to 3/4, a right triangle can be drawn with opposite side 3 and adjacent side 4, allowing the hypotenuse to be calculated. This leads to finding sin A, which can then be used in the Law of Sines formula. The final goal is to determine the length of side b, which is calculated to be approximately 11.8.
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Homework Statement


In Triangle ABC, tan A=3/4, tan B=1, and a=10. Find what b equals.


Homework Equations


You can use sina/A=sinb/B


The Attempt at a Solution


This problem is really easy using inv tangent functions and what not, but my teacher said we should be able to get it without a calculator.
Doing it with a calculator b will turn out to be 11.8. But if anyone is able to provide a detailed way to get the problem without using a calculator, that would be great.

Thanks
 
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If tanB = 1, what's the measure of angle B? That's an easy one, and one that you should know. Also, if tanA = 3/4, it's pretty easy to get sinA.
 
well the tanB=1 is equal to 45 degrees, but how can you get sinA from tanA?
 
If tanB = 1, then B is 45 degrees - that's what you meant, right?

You have tanA = 3/4. Draw a right triangle and label the side opposite to A as 3 and the side adjacent to A as 4. What does the hypotenuse have to be? From that, what's sinA?
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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