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.

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Solved Problems

Click or tap a problem to see the solution.

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:

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

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

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

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

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

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:

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

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

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

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 \(\alpha:\)

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

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:

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

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

\[\beta = \frac{\pi }{2} – \alpha .\]

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

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

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

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

\[b = 4 + 10 – \sqrt 4 = 12.\]