Before you start, make sure you know: Pythagorean theorem (Secondary Year 4 (second cycle, academic track))
Wilkin wants to know the distance between two points on the shore of Lake Enriquillo, but he can't cross the water. From where he is standing, he measures the distance to each point and the angle between the two lines of sight. How does he calculate the distance he is missing? The law of sines won't work, because he has no side with its opposite angle. That's what the law of cosines is for: a version of the Pythagorean theorem that works in any triangle.
A generalization of Pythagoras
If angle $C$ is $90^\circ$, then $\cos C=0$ and the formula becomes exactly the Pythagorean theorem. The law of cosines adds a correction term when the angle is not a right angle.
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In a triangle $ABC$, sides $b=5$ cm and $c=8$ cm and the included angle $A=60^\circ$ are known. Calculate the length of side $a$.
In a triangle $ABC$, sides $b=5$ cm and $c=8$ cm and the included angle $A=60^\circ$ are known. Calculate the length of side $a$.
In a triangle $ABC$, the sides measure $a=10$ cm and $b=12$ cm and the included angle is $C=45^\circ$. Set up the cosine rule to calculate the exact value of side $c$.
In a triangle $ABC$, the sides measure $a=10$ cm and $b=12$ cm and the included angle is $C=45^\circ$. Set up the cosine rule to calculate the exact value of side $c$.
If you know two sides and the angle between them, the third side comes from substituting directly into the formula.
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A parallelogram has two adjacent sides measuring 4 cm and 6 cm. If the included angle between them is $120^\circ$, calculate the length of the longer diagonal.
A parallelogram has two adjacent sides measuring 4 cm and 6 cm. If the included angle between them is $120^\circ$, calculate the length of the longer diagonal.
Calculate the distance between two inaccessible points $A$ and $B$, given that from an observation point $C$ the distances measured are $CA = 15$ m and $CB = 20$ m, and…
Calculate the distance between two inaccessible points $A$ and $B$, given that from an observation point $C$ the distances measured are $CA = 15$ m and $CB = 20$ m, and the visual angle $ACB$ is $40^\circ$.
If you know all three sides, solve for the cosine of the angle you want. The sign tells you whether the angle is acute (positive cosine) or obtuse (negative cosine).
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In a triangle $ABC$, the lengths of the sides are $a=3$ cm, $b=5$ cm, and $c=7$ cm. Calculate the size of angle $C$.
In a triangle $ABC$, the lengths of the sides are $a=3$ cm, $b=5$ cm, and $c=7$ cm. Calculate the size of angle $C$.
The law of cosines is useful when two distances and the angle between them can be measured, but the one you want cannot: surveying, navigation, or combining forces.
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To measure the distance across a lake between points $A$ and $B$, a surveyor stands at point $C$. He measures the distances $CA = 125.5$ m and $CB = 180.2$ m, and the…
To measure the distance across a lake between points $A$ and $B$, a surveyor stands at point $C$. He measures the distances $CA = 125.5$ m and $CB = 180.2$ m, and the angle $ACB = 75^\circ$. Calculate the distance $AB$.
In a triangle, the lengths of the sides are $x$, $x+2$, and $x+4$. If the largest angle measures $120^\circ$, set up the corresponding equation, determine the value of…
In a triangle, the lengths of the sides are $x$, $x+2$, and $x+4$. If the largest angle measures $120^\circ$, set up the corresponding equation, determine the value of $x$ and the exact lengths of the three sides.
Navigation: bearings and the distance between two vehicles
In maritime and air navigation, direction is given with a bearing: the angle that the path forms with north, measured from north in the clockwise direction. It is written with three digits. A bearing of $090^\circ$ is due east, $180^\circ$ is south, and $270^\circ$ is west.
If two vehicles leave from the same point, the angle between their paths is the difference of their bearings. That angle lies between the two distances traveled, so the law of cosines can be used. First you have to find each distance traveled: distance = speed $\times$ time.
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Two aeroplanes depart from the same airport at the same time. Aeroplane A flies at $800$ km/h on a bearing of $040^\circ$ and aeroplane B flies at $650$ km/h on a…
Two aeroplanes depart from the same airport at the same time. Aeroplane A flies at $800$ km/h on a bearing of $040^\circ$ and aeroplane B flies at $650$ km/h on a bearing of $115^\circ$. What straight-line distance separates them after 1.5 hours of flight?
The median $m_a$ is the segment that goes from vertex $A$ to the midpoint $M$ of the opposite side $a$. It is not a side of triangle $ABC$, but it is a side of triangle $ABM$, and there the law of cosines works: you know $AB=c$, $BM=\frac a2$, and angle $B$, which is the same as in the big triangle.
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In a triangle $ABC$, the sides measure $a=8$ cm, $b=10$ cm, and $c=12$ cm. Calculate the exact length of the median $m_a$ drawn from vertex $A$ to the midpoint of side…
In a triangle $ABC$, the sides measure $a=8$ cm, $b=10$ cm, and $c=12$ cm. Calculate the exact length of the median $m_a$ drawn from vertex $A$ to the midpoint of side $a$.
An inscribed (or cyclic) quadrilateral has all four vertices on a circle. These figures have a special property: opposite angles are supplementary, that is, they add up to $180^\circ$.
If $B+D=180^\circ$, then $\cos D=-\cos B$. This lets you write diagonal $AC$ in two ways, one for each triangle, and set them equal.
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A cyclic quadrilateral $ABCD$ has side lengths of $AB=3$ cm, $BC=4$ cm, $CD=5$ cm, and $DA=6$ cm. Taking into account the properties of opposite angles in a cyclic…
A cyclic quadrilateral $ABCD$ has side lengths of $AB=3$ cm, $BC=4$ cm, $CD=5$ cm, and $DA=6$ cm. Taking into account the properties of opposite angles in a cyclic quadrilateral, set up a system of equations using the cosine rule in triangles $ABC$ and $ADC$ to calculate the exact length of diagonal $AC$.
Back to the beginning: Wilkin measures $CA$, $CB$, and angle $ACB$, and applies $AB^2=CA^2+CB^2-2\cdot CA\cdot CB\cos C$ to get the distance across the lake.
Let's practice! 8 medium exercises in this section