# Thread: Need help with going through the process of solving each outer side of Right Triangle

1. ## Need help with going through the process of solving each outer side of Right Triangle

I understand how to solve all sides of a triangle using the Trig functions, ibut i'm just confused on what arithmetic to use when i'm using the Trig functions to get another side of a Right Triangle (when only one known side is given with a degree, and you are left with two unknown outsides). To make this easier here are three problems that are from my book and the process needed to solve it:

All three problems use a right triangle where the base is 'b', the opp. side is 'a', and 'c' is the hyp. The angles are 'C', 90 degrees. 'A' the point at the bottom of the base of the hyp., and 'B' the top part of the hyp.

NOTE: I'm only going to be solving for the outside not the inside.

(I couldn't get latex to print this out correctly, i hope this is readable to the viewer.)

1)

A = 23.5 b = 10
10 tan(23.5) = 4.34 (a)

10 cos(23.5) = 10.90 (c)

2)
A = 52.6 c = 54

54 sin 52.6 = 42.90 (a)

54 cos 52.6

3)

B 16.8 b = 30

30.5 / tan(16.8) = 101.02

30.5 / sin (16.8) = 105.52

** My question is, when i'm solving for this how do i know when to divide the angle inside the trig function, -for example 10/cos(23.5) - by the given outside length inside of multiplying the length by the trig function -for example 54 sin (52.6)- ?

2. Originally Posted by MajorJohnson
I understand how to solve all sides of a triangle using the Trig functions, ibut i'm just confused on what arithmetic to use when i'm using the Trig functions to get another side of a Right Triangle (when only one known side is given with a degree, and you are left with two unknown outsides). To make this easier here are three problems that are from my book and the process needed to solve it:

All three problems use a right triangle where the base is 'b', the opp. side is 'a', and 'c' is the hyp. The angles are 'C', 90 degrees. 'A' the point at the bottom of the base of the hyp., and 'B' the top part of the hyp.

NOTE: I'm only going to be solving for the outside not the inside.

(I couldn't get latex to print this out correctly, i hope this is readable to the viewer.)

1)

A = 23.5 b = 10
10 tan(23.5) = 4.34 (a)

10 cos(23.5) = 10.90 (c)

2)
A = 52.6 c = 54

54 sin 52.6 = 42.90 (a)

54 cos 52.6

3)

B 16.8 b = 30

30.5 / tan(16.8) = 101.02

30.5 / sin (16.8) = 105.52

** My question is, when i'm solving for this how do i know when to divide the angle inside the trig function, -for example 10/cos(23.5) - by the given outside length inside of multiplying the length by the trig function -for example 54 sin (52.6)- ?
When you learn the trig functions, you should learn something like
sine= opposite side/hypotenuse
cosine= near side/hypotenuse
tangent= near side/opposite side

And when you are given one side or another of a right triangle and an angle, write what you are given in exactly that way.

If, as in (1) you are given angle A and side b (and you are using the convention that "side a" is opposite to "angle A") to find a, you want to find the opposite side and you know the near side. The formula that involves both "opposite side" and "near side" is "tangent= near side/opposite side" so, using the information you are given:
$\displaystyle tan(23.5)= \frac{a}{10}$

Now, you should have learned enough algebra to know that to solve that equation for a, you have to multiply both sides by 10: $\displaystyle 10(tan(23.5)= 10\frac{a}{10}= a$

To find the length of the hypotenuse, realize that the fraction that involves "near side" and "hypotenuse" is "cosine= near side/hypotenuse" which, using the data given becomes $\displaystyle cos(23.5)= \frac{10}{c}$

. Now, solve that for c. That's just little harder because now the value you want, c, is in the denominator. Multiply both sides by c and you get $\displaystyle c cos(23.5)= \frac{10}{c}c= 10$. Finally, to get c, divide both sides by cos(23.5): $\displaystyle c cos(23.5)/cos(23.5)= c= \frac{10}{cos(23.5)}$.

### A=23.5 B= 10

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