How to find x in a triangle

There are many ways to find the side length of a right triangle. We are going to focus on two specific cases.

Table of Contents Show

Video Tutorial on Finding the Side Length of a Right Triangle
Practice Problems
Another Method
Largest Angle?

Case II

We know 1 side and 1 angle of the right triangle, in which case, use sohcahtoa.

Video Tutorial on Finding the Side Length of a Right Triangle

Practice Problems

Calculate the length of the sides below. In each case, round your answer to the nearest hundredth.

Problem 1

Find the length of side X in the triangle below.

Step 2

Substitute the two known sides into the Pythagorean theorem's formula:

$$ a^2 + b^2 = c^2 \\ 8^2 + 6^2 = x^2 \\ 100 = x^2 \\ x = \sqrt{100} \\ x = \boxed{10} $$

Problem 2

Find the length of side X in the right triangle below.

Step 1

Since we know 1 side and 1 angle of this triangle, we will use sohcahtoa.

Step 2

Set up an equation using a sohcahtoa ratio. Since we know the hypotenuse and want to find the side opposite of the 53° angle, we are dealing with sine

$$ sin(53) = \frac{ opposite}{hypotenuse} \\ sin(53) = \frac{ \red x }{ 12 } $$

Now, just solve the Equation:

Step 3

$$ sin(53) = \frac{ \red x }{ 12 } \\ \red x = 12 \cdot sin (53) \\ \red x = \boxed{ 11.98} $$

Problem 3

Find the length of side X in the right triangle below.

Step 2

Substitute the two known sides into the Pythagorean theorem's formula:

$$ a^2 + b^2 = c^2 \\ \red t^2 + 12^2 = 13^2 \\ \red t^2 + 144 = 169 \\ \red t^2 = 169 - 144 \\ \red t^2 = 25 \\ \red t = \boxed{5} $$

Problem 4

Find the length of side X in the right triangle below.

Step 1

Since we know 1 side and 1 angle of this triangle, we will use sohcahtoa.

Step 2

Set up an equation using the sine, cosine or tangent ratio Since we want to know the length of the hypotenuse, and we already know the side opposite of the 53° angle, we are dealing with sine.

$$ sin(67) = \frac{opp}{hyp} \\ sin(67) = \frac{24}{\red x} $$

Now, just solve the Equation:

Step 3

$$ x = \frac{ 24}{ sin(67) } \\ x = 26.07 $$

Problem 5

Calculate the length of side X in the right triangle below.

Step 1

Since we know 2 sides and 1 angle of this triangle, we can use either the Pythagorean theorem (by making use of the two sides) or use sohcahtoa (by making use of the angle and 1 of the given sides).

Step 2

Chose which way you want to solve this problem. There are several different solutions. The only thing you cannot use is sine, since the sine ratio does not involve the adjacent side, x, which we are trying to find.

The answers are slightly different (tangent s 35.34 vs 36 for the others) due to rounding issues. I rounded the angle's measure to 23° for the sake of simplicity of the diagram. A more accurate angle measure would have been 22.61986495°. If you use that value instead of 23°, you will get answers that are more consistent.

Step 3

$$ x = \frac{ 24}{ sin(67) } \approx 26.07 $$

"SSS" means "Side, Side, Side"

"SSS" is when we know three sides of the triangle, and want to find the missing angles.

We use the "angle" version of the Law of Cosines:

cos(C) = a2 + b2 − c2 2ab

cos(A) = b2 + c2 − a2 2bc

cos(B) = c2 + a2 − b2 2ca

(they are all the same formula, just different labels)

Example 1

In this triangle we know the three sides:

a = 8,
b = 6 and
c = 7.

Use The Law of Cosines first to find one of the angles. It doesn't matter which one. Let's find angle A first:

cos A = (b2 + c2 − a2) / 2bc

cos A = (62 + 72 − 82) / (2×6×7)

cos A = (36 + 49 − 64) / 84

cos A = 0.25

A = cos−1(0.25)

A = 75.5224...°

A = 75.5° to one decimal place.

Next we will find another side. We use The Law of Cosines again, this time for angle B:

cos B = (c2 + a2 − b2)/2ca

cos B = (72 + 82 − 62)/(2×7×8)

cos B = (49 + 64 − 36) / 112

cos B = 0.6875

B = cos−1(0.6875)

B = 46.5674...°

B = 46.6° to one decimal place

Finally, we can find angle C by using 'angles of a triangle add to 180°':

C = 180° − 75.5224...° − 46.5674...°

C = 57.9° to one decimal place

Now we have completely solved the triangle i.e. we have found all its angles.

The triangle can have letters other than ABC:

Example 2

This is also an SSS triangle.

In this triangle we know the three sides x = 5.1, y = 7.9 and z = 3.5. Use The Law of Cosines to find angle X first:

cos X = (y2 + z2 − x2)/2yz

cos X = ((7.9)2 + (3.5)2 − (5.1)2)/(2×7.9×3.5)

cos X = (62.41 + 12.25 − 26.01)/55.3

cos X = 48.65/55.3 = 0.8797...

X = cos−1(0.8797...)

X = 28.3881...°

X = 28.4° to one decimal place

Next we will use The Law of Cosines again to find angle Y:

cos Y = (z2 + x2 − y2)/2zx

cos Y = −24.15/35.7 = −0.6764...

cos Y = (12.25 + 26.01 − 62.41)/35.7

cos Y = −24.15/35.7 = −0.6764...

Y = cos−1(−0.6764...)

Y = 132.5684...°

Y = 132.6° to one decimal place.

Finally, we can find angle Z by using 'angles of a triangle add to 180°':

Z = 180° − 28.3881...° − 132.5684...°

Z = 19.0° to one decimal place

Another Method

Largest Angle?

Why do we try to find the largest angle first? That way the other two angles must be acute (less than 90°) and the Law of Sines will give correct answers.

The Law of Sines is difficult to use with angles above 90°. There can be two answers either side of 90° (example: 95° and 85°), but a calculator will only give you the smaller one.

So by calculating the largest angle first using the Law of Cosines, the other angles are less than 90° and the Law of Sines can be used on either of them without difficulty.