# 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 $$\left({SSS}\right),$$ or by a side and two adjacent angles $$\left({ASA}\right).$$

In a right triangle, one of the angles is equal to $$90^\circ.$$ 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 $$\left({HL}\right),$$
• Two legs $$\left({LL}\right),$$
• The hypotenuse and an angle $$\left({HA}\right),$$
• A leg and the adjacent angle $$\left({LA_A}\right),$$
• A leg and the opposite angle $$\left({LA_O}\right),$$

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,

${\sin \alpha = \frac{a}{c}, \;\text{ or }\; }\kern0pt{\cos \alpha = \frac{b}{c}.}$

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

${\cos \beta = \frac{a}{c}, \;\text{ or }\; }\kern0pt{\sin \beta = \frac{b}{c}.}$

### $$2.$$ 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:

${\sin \alpha = \frac{a}{c}, \;\text{ or }\; }\kern0pt{\cos \alpha = \frac{b}{c},}$

${\cos \beta = \frac{a}{c}, \;\text{ or }\; }\kern0pt{\sin \beta = \frac{b}{c}.}$

### $$3.$$ 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:

${a = c\sin \alpha ,\;\;}\kern0pt{b = c\cos \alpha .}$

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.

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

A right triangle has a hypotenuse of length $$41$$ and a leg of length $$9.$$ Find the other leg and two acute angles.

### Example 2

The two legs of a right triangle are $$12$$ and $$35.$$ Find the hypotenuse and two acute angles.

### Example 3

In a right triangle, the length of the hypotenuse is $$20$$ and an acute angle is $$30^\circ.$$ Find the missing elements.

### Example 4

In a right triangle, one angle is $$50^\circ$$ and the adjacent leg is $$15.$$ Find the missing elements.

### Example 5

A leg in a right triangle is equal to $$25,$$ and the angle opposite to this leg is equal to $$30^\circ.$$ Determine the missing elements.

### Example 6

Let $$a$$ be a leg of a right triangle and $$h$$ be the altitude drawn to the hypotenuse. Determine the missing elements of the triangle.

### Example 7

The perimeter of a right triangle is $$P$$ and its area is $$A.$$ Find the length of the hypotenuse.

### Example 8

A right triangle with an angle of $$\alpha$$ has an area $$A.$$ Find the missing elements of the triangle.

### Example 9

In a right triangle, the projections of the legs on the hypotenuse are $$m = 9$$ and $$n = 16.$$ Find the radius of the inscribed circle.

### Example 10

The radii of the circumcircle and incircle of a right triangle are $$R$$ and $$r.$$ Find the legs of the triangle.

### Example 1.

A right triangle has a hypotenuse of length $$41$$ and a leg of length $$9.$$ Find the other leg and two acute angles.

Solution.

According to the notation above, we deal here with an $$HL$$ model. Denote $$c = 41$$ and $$a = 9.$$ The acute angle $$\alpha$$ is supposed to be opposite to side $$a.$$

We find the $$b$$ using the Pythagorean theorem:

${b = \sqrt {{c^2} – {a^2}} }={ \sqrt {{{41}^2} – {9^2}} }={ \sqrt {1681 – 81} }={ \sqrt {1600} }={ 40.}$

The angles $$\alpha$$ and $$\beta$$ can be specified by the sine function:

${\sin \alpha = \frac{a}{c} = \frac{9}{{41}},\;\;}\kern0pt{\sin \beta = \frac{b}{c} = \frac{{40}}{{41}}.}$

### Example 2.

The two legs of a right triangle are $$12$$ and $$35.$$ Find the hypotenuse and two acute angles.

Solution.

The problem is classified as a $$LL$$ model. Let $$a = 12$$ and $$b = 35.$$ The length of the hypotenuse is given by the Pythagorean theorem:

${c = \sqrt {{a^2} + {b^2}} }={ \sqrt {{{12}^2} + {{35}^2}} }={ \sqrt {144 + 1225} }={ \sqrt {1369} }={ 37.}$

Assuming that the angle $$\alpha$$ lies opposite the side $$a,$$ we have

${\sin \alpha = \frac{a}{c} = \frac{12}{{37}},\;\;}\kern0pt{\sin \beta = \frac{b}{c} = \frac{{35}}{{37}}.}$

### Example 3.

In a right triangle, the length of the hypotenuse is $$20$$ and an acute angle is $$30^\circ.$$ Find the missing elements.

Solution.

Here we have an $$HA$$ scenario. We denote the hypotenuse by $$c$$ and the known angle by $$\alpha.$$ Using the trigonometric functions, we obtain

${a = c\sin \alpha }={ 20\sin {30^\circ} }={ 20 \times \frac{1}{2} }={ 10;}$

${b = c\cos \alpha }={ 20\cos {30^\circ}.}$

If $$\sin 30^\circ = \large{\frac{1}{2}}\normalsize,$$ then

${\cos {30^\circ} = \sqrt {1 – {{\sin }^2}{{30}^\circ}} }={ \sqrt {1 – {{\left( {\frac{1}{2}} \right)}^2}} }={ \sqrt {1 – \frac{1}{4}} }={ \sqrt {\frac{3}{4}} }={ \frac{{\sqrt 3 }}{2}.}$

So the length of the second leg $$b$$ is given by

${b = 20\cos {30^\circ} }={ 20 \times \frac{{\sqrt 3 }}{2} }={ 10\sqrt 3 .}$

The complementary angle $$\beta$$ is equal to

${\beta = {90^\circ} – \alpha }={ {90^\circ} – {30^\circ} }={ {60^\circ}.}$

### Example 4.

In a right triangle, one angle is $$50^\circ$$ and the adjacent leg is $$15.$$ Find the missing elements.

Solution.

This is a $$LA_A$$ model by the above notation. The other acute angle is obviously equal to

${\beta = {90^\circ} – \alpha }={ {90^\circ} – {50^\circ} }={ {40^\circ}.}$

If $$a$$ is the leg opposite to the angle $$\alpha,$$ we can write

${c = \frac{a}{{\sin \alpha }} }={ \frac{{15}}{{\sin {{50}^\circ}}}.}$

Using a calculator, we find the value of $$\sin 50^\circ:$$

$\sin {50^\circ} = 0.7660$

Hence,

${c = \frac{{15}}{{\sin {{50}^0}}} }={ \frac{{15}}{{0.7660}} }\approx{ 19.58}$

The second leg can be found by the Pythagorean theorem:

${b = \sqrt {{c^2} – {a^2}} }={ \sqrt {{{19.58}^2} – {{15}^2}} }={ \sqrt {158.38} }\approx{ 12.58}$

### Example 5.

A leg in a right triangle is equal to $$25,$$ and the angle opposite to this leg is equal to $$30^\circ.$$ Determine the missing elements.

Solution.

This triangle matches to the $$LA_O$$ definition. Suppose the known leg is $$a$$ and the opposite angle is $$\alpha.$$ Then, the hypotenuse is given by

${c = \frac{a}{{\sin \alpha }} = \frac{{25}}{{\sin {{30}^\circ}}} }={ \frac{{25}}{{\frac{1}{2}}} }={ 50.}$

There are several ways to find the other leg. Let’s use the Pythagorean theorem:

${b = \sqrt {{c^2} – {a^2}} }={ \sqrt {{{50}^2} – {{25}^2}} }={ \sqrt {2500 – 625} }={ \sqrt {1875} }={ 25\sqrt 3 .}$

The complementary angle is

${\beta = {90^\circ} – \alpha }={ {90^\circ} – {30^\circ} }={ {60^\circ}.}$

### Example 6.

Let $$a$$ be a leg of a right triangle and $$h$$ be the altitude drawn to the hypotenuse. Determine the missing elements of the triangle.

Solution.

For the inner triangle $$BDC,$$ the following relationship is valid:

${h = a\sin \beta ,}\;\; \Rightarrow {\sin \beta = \frac{h}{a}.}$

The cosine of $$\beta$$ can be found using the Pythagorean trig identity:

${\cos \beta = \sqrt {1 – {{\sin }^2}\beta } }={ \sqrt {1 – {{\left( {\frac{h}{a}} \right)}^2}} }={ \sqrt {\frac{{{a^2} – {h^2}}}{{{a^2}}}} }={ \frac{{\sqrt {{a^2} – {h^2}} }}{a}.}$

Now we can find the hypotenuse $$c:$$

${c = \frac{a}{{\cos \beta }} }={ \frac{a}{{\frac{{\sqrt {{a^2} – {h^2}} }}{a}}} }={ \frac{{{a^2}}}{{\sqrt {{a^2} – {h^2}} }}.}$

The other leg $$b$$ is given by

${b = c\sin \beta }={ \frac{{{a^2}}}{{\sqrt {{a^2} – {h^2}} }} \times \frac{h}{a} }={ \frac{{ah}}{{\sqrt {{a^2} – {h^2}} }}.}$

Finally, calculate the cosine of $$\alpha:$$

${\cos \alpha = \frac{b}{c} }={ \frac{{\frac{{ah}}{{\sqrt {{a^2} – {h^2}} }}}}{{\frac{{{a^2}}}{{\sqrt {{a^2} – {h^2}} }}}} }={ \frac{h}{a}.}$

### Example 7.

The perimeter of a right triangle is $$P$$ and its area is $$A.$$ Find the length of the hypotenuse.

Solution.

We denote the legs by $$a, b$$ and the hypotenuse by $$c.$$ By condition,

${a + b + c = P,}\;\; \Rightarrow {a + b = P – c.}$

The area is given by the formula

${A = \frac{{ab}}{2},}\;\; \Rightarrow {2ab = 4A.}$

Consider the sum of the legs squared:

${{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2} }={ {c^2} + 4A,}$

where we used the Pythagorean theorem: $${a^2} + {b^2} = {c^2}.$$

From the other side, we have

$a + b = P – c.$

Therefore, we get the following equation for $$c:$$

${\left( {P – c} \right)^2} = {c^2} + 4A.$

It has a simple solution:

$\require{cancel}{{P^2} – 2Pc + \cancel{c^2} = \cancel{c^2} + 4A,}\;\; \Rightarrow {2Pc = {P^2} – 4A,}\;\; \Rightarrow {c = \frac{{{P^2} – 4A}}{{2P}}.}$

### Example 8.

A right triangle with an angle of $$\alpha$$ has an area $$A.$$ Find the missing elements of the triangle.

Solution.

The area of a right triangle is given by the formula

$A = \frac{{ab}}{2},$

where $$a$$ and $$b$$ are the legs of the triangle.

Let $$c$$ be the (unknown) hypotenuse of the triangle. The legs can be expressed in terms of $$c$$ and $$\alpha:$$

${a = c\sin \alpha ,\;\;}\kern0pt{b = c\cos \alpha .}$

We assume here that the angle $$\alpha$$ is opposite to the side $$a.$$

Then we have

${A = \frac{{ab}}{2} }={ \frac{1}{2}{c^2}\sin \alpha \cos \alpha .}$

Solve this equation for $$c:$$

${{c^2} = \frac{{2A}}{{\sin \alpha \cos \alpha }},}\;\; \Rightarrow {c = \sqrt {\frac{{2A}}{{\sin \alpha \cos \alpha }}} .}$

We now derive the expressions for legs:

$\require{cancel}{a = c\sin \alpha }={ \sqrt {\frac{{2A}}{{\sin \alpha \cos \alpha }}} \times \sin \alpha }={ \sqrt {\frac{{2A{{\sin }^{\cancel{2}}}\alpha }}{{\cancel{\sin \alpha} \cos \alpha }}} }={ \sqrt {\frac{{2A\sin \alpha }}{{\cos \alpha }}} }={ \sqrt {2A\tan \alpha } ;}$

${b = c\cos \alpha }={ \sqrt {\frac{{2A}}{{\sin \alpha \cos \alpha }}} \times \cos \alpha }={ \sqrt {\frac{{2A{{\cos }^{\cancel{2}}}\alpha }}{{\sin \alpha \cancel{\cos \alpha }}}} }={ \sqrt {\frac{{2A\cos \alpha }}{{\sin \alpha }}} }={ \sqrt {2A\cot \alpha } .}$

The complentary angle is obviously equal to

$\beta = \frac{\pi }{2} – \alpha .$

### Example 9.

In a right triangle, the projections of the legs on the hypotenuse are $$m = 9$$ and $$n = 16.$$ Find the radius of the inscribed circle.

Solution.

Applying the Pythagorean theorem to the triangles $$ADC$$ and $$BDC,$$ we get

${{b^2} – {m^2} = {b^2} – {9^2} = {h^2},\;\;}\kern0pt{{a^2} – {n^2} = {a^2} – {16^2} = {h^2}.}$

Hence,

${{b^2} – 81 = {a^2} – 256, \;\;}\kern0pt{\text{or }\; {b^2} = {a^2} – 175.}$

The second equation is given by Pythagorean theorem for the triangle $$ABC:$$

${{a^2} + {b^2} = {c^2} }={ {\left( {m + n} \right)^2} }={ {25^2} }={ 625.}$

Thus, we have a system of two equations with the variables $$a^2$$ and $$b^2:$$

$\left\{ \begin{array}{l} {b^2} = {a^2} – 175\\ {a^2} + {b^2} = 625 \end{array} \right..$

Solving the system yields the following results:

${{a^2} + {a^2} – 175 = 625,}\;\; \Rightarrow {2{a^2} = 800,}\;\; \Rightarrow {{a^2} = 400,}\;\; \Rightarrow {a = 20;}$

${{b^2} = {a^2} – 175 }={ 400 – 175 }={ 225,}\;\; \Rightarrow {b = 15.}$

The radius of the inscribed circle is given by

${r = \frac{{a + b – c}}{2} }={ \frac{{20 + 15 – 25}}{2} }={ 5.}$

### Example 10.

The radii of the circumcircle and incircle of a right triangle are $$R$$ and $$r.$$ Find the legs of the triangle.

Solution.

Let $$AB = c$$ is the hypotenuse and $$BC = a,$$ $$AC = b$$ are the legs. The figure $$CKOL$$ is a square with side $$r.$$ Suppose that the length of the line segment $$AK$$ is $$x.$$ The secant segments $$AK$$ and $$AM$$ have the same length $$x.$$

Recall that the hypotenuse length is twice the radius of the circumcircle: $$c = 2R.$$ The secant segments $$BM$$ and $$BL$$ have the same length equal to $$2R – x.$$

We express the sides of the triangle in terms of $$x$$ and the radii $$R$$ and $$r:$$

${a = AC = r + x,\;\;}\kern0pt{b = BC = r + 2R – x,\;\;}\kern0pt{c = AB = 2R.}$

By the Pythagorean theorem,

${{a^2} + {b^2} = {c^2},}\;\; \Rightarrow {{\left( {r + x} \right)^2} + {\left( {r + 2R – x} \right)^2} = 4{R^2}.}$

Now we need to solve the equation for $$x.$$ To simplify it, we use the identities for the square of a sum and the square of a trinomial:

${\left( {x + y} \right)^2} = {x^2} + 2xy + {y^2},$

${{\left( {x + y + z} \right)^2} }={ {x^2} + {y^2} + {z^2} }+{ 2xy + 2xz + 2yz.}$

Hence,

${{\left( {r + x} \right)^2} + {\left( {r + 2R – x} \right)^2} }={ 4{R^2},}$

$\Rightarrow {{r^2} + \cancel{2rx} + {x^2} }+{ {r^2} + \cancel{4{R^2}} + {x^2} }+{ 4rR – \cancel{2rx} – 4Rx }={ \cancel{4{R^2}},}$

$\Rightarrow 2{x^2} – 4Rx + 2{r^2} + 4rR = 0,$

$\Rightarrow {x^2} – 2Rx + {r^2} + 2rR = 0.$

This quadratic equation has two solutions:

${{x_{1,2}} }={ \frac{{2R \pm \sqrt {4{R^2} – 4{r^2} – 8rR} }}{2} }={ R \pm \sqrt {{R^2} – {r^2} – 2rR} .}$

If we take the first root

${x_1} = R – \sqrt {{R^2} – {r^2} – 2rR},$

we get the following lengths of the legs:

${a = r + x }={ r + R – \sqrt {{R^2} – {r^2} – 2rR} ,}$

${b = r + 2R – x }={ r + 2R – R }+{ \sqrt {{R^2} – {r^2} – 2rR} }={ r + R + \sqrt {{R^2} – {r^2} – 2rR} .}$

It is clear that the second root $$x_2$$ produces the same legs only rearranged in places. Thus, the final answer is

${a = r + R – \sqrt {{R^2} – {r^2} – 2rR} ,}$

$b = r + R + \sqrt {{R^2} – {r^2} – 2rR} .$

Let’s substitute some sample values. If $$R = 10$$ and $$r = 4,$$ we obtain

${a = 4 + 10 }-{ \sqrt {{{10}^2} – {4^2} – 2 \cdot 4 \cdot 10} }={ 4 + 10 + \sqrt 4 }={ 16,}$

$b = 4 + 10 – \sqrt 4 = 12.$