# Precalculus

## Trigonometry # Solving Right Triangles

An arbitrary triangle is defined by three components, at least one of which must be a side. For example, a triangle can be defined by three sides (SSS), or by a side and two adjacent angles (ASA).

In a right triangle, one of the angles is equal to 90°. Therefore, to find all sides and angles of a right triangle, it is enough to know only two elements, at least one of which must be a side.

There are 5 basic combinations of sides and angles that uniquely define a right triangle:

• The hypotenuse and a leg (HL),
• Two legs (LL),
• The hypotenuse and an angle (HA)
• A leg and the adjacent angle (LAA),
• A leg and the opposite angle (LAO).

Solving a triangle means finding all its sides and angles. Under finding an angle we mean finding a trigonometric function of the angle.

Depending on the given data, we use different relationships and identities. Let's consider these scenarios in more detail.

## 1. Solving a Right Triangle Given the Hypotenuse and a Leg $$\left({HL}\right)$$

Suppose the hypotenuse $$c$$ and leg $$a$$ are known for a right triangle.

To find the other leg $$b,$$ we use the Pythagorean theorem:

${b^2} = {c^2} - {a^2},\;\; \Rightarrow b = \sqrt {{c^2} - {a^2}} .$

Since all $$3$$ sides are already known, we can determine any angle using a trigonometric function. For example,

$\sin \alpha = \frac{a}{c}, \;\text{ or }\; \cos \alpha = \frac{b}{c}.$

The other acute angle can be expressed in terms of the sine or cosine as

$\cos \beta = \frac{a}{c}, \;\text{ or }\; \sin \beta = \frac{b}{c}.$

## 2. Solving a Right Triangle Given Two Legs $$\left({LL}\right)$$

Let a right triangle be defined by legs $$a$$ and $$b.$$

The hypotenuse $$c$$ can be easily found by the Pythagorean theorem:

${c^2} = {a^2} + {b^2},\;\; \Rightarrow c = \sqrt {{a^2} + {b^2}} .$

Now we are at the same point as in the previous case, so the acute angles $$\alpha$$ and $$\beta$$ can be defined in the same way:

$\sin \alpha = \frac{a}{c}, \;\text{ or }\; \cos \alpha = \frac{b}{c},$
$\cos \beta = \frac{a}{c}, \;\text{ or }\; \sin \beta = \frac{b}{c}.$

## 3. Solving a Right Triangle Given the Hypotenuse and an Angle $$\left({HA}\right)$$

A right triangle has a hypotenuse $$c$$ and an angle $$\alpha.$$

We find the legs of the triangle using the Pythagorean trig identity:

$a = c\sin \alpha ,\;\;b = c\cos \alpha .$

The other acute angle of the triangle is obviously

$\beta = {90^\circ} - \alpha .$

## 4. Solving a Right Triangle Given a Leg and the Adjacent Angle $$\left({LA_A}\right)$$

Consider a right triangle and suppose its leg $$b$$ and the adjacent angle $$\alpha$$ are known.

The hypotenuse can be found by the formula

$c = \frac{b}{{\cos \alpha }}.$

The other leg $$a$$ is given by

$a = b\tan \alpha .$

The angles $$\alpha$$ and $$\beta$$ are complementary, so

$\beta = {90^\circ} - \alpha .$

## 5. Solving a Right Triangle Given a Leg and the Opposite Angle $$\left({LA_O}\right)$$

The leg $$a$$ and the opposite angle $$\alpha$$ of a right triangle are known.

In this case, the hypotenuse can be found by the formula

$c = \frac{a}{{\sin \alpha }}.$

To determine the other leg $$b,$$ we can use the identity

$b = a\cot \alpha .$

The angle $$\beta$$ is complementary to $$\alpha:$$

$\beta = {90^\circ} - \alpha .$

## General Case: Solving a Right Triangle Given Two Arbitrary Elements

In general, a right triangle can be defined by two arbitrary distinct elements. One of them must be a metric element (like a leg). For example, we could define a right triangle by its area and perimeter, or say, by an altitude drawn to the hypotenuse and a radius of the inscribed circle. Some pairs of such parameters lead to complex systems of equations. However, there is also a plenty of combinations of this type which are quite solvable. We will consider some of these problems below.

See solved problems on Page 2.