Solving triangles

Law of sines, law of cosines, law of tangents and the area of a triangle. 9 problems with step-by-step solutions.

Solve each problem in your notebook first, then open the solution. In all answers \(k, m \in \mathbb{Z}\).

Problem 1 Law of cosines

A triangle has sides \(a = 7\) cm, \(b = 5\) cm and \(m(\angle C) = 60^\circ\). Find side \(c\).

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\[c^2 = a^2 + b^2 - 2ab\cos C = 49 + 25 - 2\cdot 7\cdot 5\cdot\frac12 = 39.\]

\(c = \sqrt{39} \approx 6.24\) cm.

Problem 2 Law of sines

In triangle \(ABC\), \(a = 8\) cm, \(m(\angle A) = 45^\circ\) and \(m(\angle C) = 105^\circ\). Find side \(b\).

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\(m(\angle B) = 180^\circ - 45^\circ - 105^\circ = 30^\circ\).

\[\frac{a}{\sin A} = \frac{b}{\sin B} \Rightarrow b = \frac{8\sin 30^\circ}{\sin 45^\circ} = \frac{8\cdot\frac12}{\frac{\sqrt2}{2}} = \frac{8}{\sqrt2} = 4\sqrt2 \text{ cm}.\]

Problem 3 Inscribed and circumscribed circle

The sides of a triangle are \(a = 7\) cm, \(b = 8\) cm and \(c = 9\) cm. Find its area and the radii \(r\) and \(R\) of the inscribed and circumscribed circles.

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Half the perimeter: \(u = \frac{7 + 8 + 9}{2} = 12\).

Heron's formula: \(A = \sqrt{12\cdot 5\cdot 4\cdot 3} = \sqrt{720} = 12\sqrt5\) cm².

\(A = u\cdot r \Rightarrow r = \frac{12\sqrt5}{12} = \sqrt5\) cm.

\(A = \frac{abc}{4R} \Rightarrow R = \frac{7\cdot 8\cdot 9}{4\cdot 12\sqrt5} = \frac{504}{48\sqrt5} = \frac{21}{2\sqrt5} = \frac{21\sqrt5}{10}\) cm.

Problem 4 Two triangles with a common vertex

The segments \(AE\) and \(BD\) intersect at \(C\), so that \(ABC\) and \(DCE\) are triangles. \(AB = 7\), \(AC = 5\), \(BC = 3\), \(CD = 4\) and \(CE = 6\). Find \(DE\).

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The angles \(\angle ACB\) and \(\angle DCE\) are vertical angles, so they are equal. Call this angle \(\gamma\).

In triangle \(ABC\): \(\cos\gamma = \frac{AC^2 + BC^2 - AB^2}{2\cdot AC\cdot BC} = \frac{25 + 9 - 49}{30} = -\frac12\), so \(\gamma = 120^\circ\).

In triangle \(DCE\): \(DE^2 = 16 + 36 - 2\cdot 4\cdot 6\cdot\left(-\frac12\right) = 52 + 24 = 76\), so \(DE = \sqrt{76} = 2\sqrt{19}\).

Problem 5 Side and circumradius

In triangle \(ABC\), \(AC = 6\), \(m(\angle A) = 30^\circ\) and \(m(\angle C) = 105^\circ\). Find \(BC = a\) and the radius \(R\) of the circumscribed circle.

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\(m(\angle B) = 180^\circ - 30^\circ - 105^\circ = 45^\circ\), and \(AC = b = 6\).

\(2R = \frac{b}{\sin B} = \frac{6}{\frac{\sqrt2}{2}} = 6\sqrt2\), so \(R = 3\sqrt2\).

\(a = 2R\sin A = 6\sqrt2\cdot\frac12 = 3\sqrt2\).

Problem 6 Area of a triangle

a) Two sides of a triangle are \(a = 5\) cm and \(b = 8\) cm, and \(m(\angle C) = 30^\circ\). Find the area.

b) Two sides of a triangle are 4 cm and 6 cm, and the angle between them is \(135^\circ\). Find the area.

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Use \(A = \frac12 ab\sin C\).

a) \(A = \frac12\cdot 5\cdot 8\cdot\frac12 = 10\) cm².

b) \(\sin 135^\circ = \frac{\sqrt2}{2}\), so \(A = \frac12\cdot 4\cdot 6\cdot\frac{\sqrt2}{2} = 6\sqrt2\) cm².

Problem 7 Angle in a square

\(ABCD\) is a square and \(E\) is the midpoint of side \(CD\). Let \(x = m(\angle EAC)\). Find \(\sin x\).

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Put the square in a coordinate system with side 2: \(A(0,0)\), \(B(2,0)\), \(C(2,2)\), \(D(0,2)\), so \(E(1,2)\).

The ray \(AC\) makes the angle \(45^\circ\) with the \(x\)-axis (\(\tan = 1\)). The ray \(AE\) makes an angle \(\beta\) with \(\tan\beta = 2\). So \(x = \beta - 45^\circ\):

\[\tan x = \frac{2 - 1}{1 + 2\cdot 1} = \frac13.\]

Then \(\sin x = \frac{1}{\sqrt{1 + 9}} = \frac{1}{\sqrt{10}} = \frac{\sqrt{10}}{10}\).

Problem 8 Finding an angle

a) The sides of a triangle are 3, 5 and 7. Find its largest angle.

b) In triangle \(ABC\), \(c^2 = a^2 + b^2 - \sqrt3\,ab\). Find \(m(\angle C)\).

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a) The largest angle is opposite the longest side 7: \(\cos\gamma = \frac{3^2 + 5^2 - 7^2}{2\cdot 3\cdot 5} = \frac{-15}{30} = -\frac12\), so \(\gamma = 120^\circ\).

b) Compare with \(c^2 = a^2 + b^2 - 2ab\cos C\): \(2\cos C = \sqrt3\), so \(\cos C = \frac{\sqrt3}{2}\) and \(m(\angle C) = 30^\circ\).

Problem 9 Law of tangents

a) Show that in any triangle \(\dfrac{\sin A + \sin B}{\sin A - \sin B} = \dfrac{a + b}{a - b}\) (for \(a \ne b\)).

b) In triangle \(ABC\), \(a = 5\), \(b = 3\) and \(m(\angle C) = 60^\circ\). Use the law of tangents to find the angles \(A\) and \(B\).

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a) By the law of sines \(\sin A = \frac{a}{2R}\) and \(\sin B = \frac{b}{2R}\). Then \(\frac{\sin A + \sin B}{\sin A - \sin B} = \frac{(a + b)/2R}{(a - b)/2R} = \frac{a + b}{a - b}\).

b) \(A + B = 120^\circ\). The law of tangents: \(\frac{a + b}{a - b} = \frac{\tan\frac{A + B}{2}}{\tan\frac{A - B}{2}}\), so \(4 = \frac{\tan 60^\circ}{\tan\frac{A - B}{2}}\) and \(\tan\frac{A - B}{2} = \frac{\sqrt3}{4}\).

\(\frac{A - B}{2} \approx 23.41^\circ\). With \(\frac{A + B}{2} = 60^\circ\): \(A \approx 83.41^\circ\), \(B \approx 36.59^\circ\).

Check with the law of cosines: \(c^2 = 25 + 9 - 15 = 19\), \(\cos A = \frac{9 + 19 - 25}{2\cdot 3\sqrt{19}} \approx 0.115\), \(A \approx 83.4^\circ\).