Solving non-right-angled triangles
Two Dublin Bus routes leave the same depot at Phibsborough, 75° apart, and in one hour one bus covers 12 km and the other 16 km (taking both routes as straight lines). You want to know how far apart the buses are after that hour. There is no right angle anywhere. You already know the sine rule and the cosine rule separately. Now, given any triangle with three pieces of data, you must decide which rule to apply first and solve it completely. We will come back to the buses (and two ships out of Rosslare) later.
What data you have, which rule to use
Three pieces of data that include at least one side can allow you to solve a triangle, but they must be compatible and do not always determine a unique triangle. In the side-side-angle (non-included angle) case there may be zero, one or two solutions (see the ambiguous case below). Depending on the data, the shortest route is different.
In a triangle $ABC$, the sides $a = 5$ cm and $b = 8$ cm are known, as well as the included angle $C = 60^\circ$ between them. Calculate the length of side $c$.
In a triangle $ABC$, the sides $a = 5$ cm and $b = 8$ cm are known, as well as the included angle $C = 60^\circ$ between them. Calculate the length of side $c$.
The three sides of a triangle measure $a = 3$ m, $b = 5$ m and $c = 7$ m. Determine the size of the largest angle in this triangle.
The three sides of a triangle measure $a = 3$ m, $b = 5$ m and $c = 7$ m. Determine the size of the largest angle in this triangle.
In a triangle, the following data are known: $a = 12$ m, $b = 8$ m and angle $A = 120^\circ$. Calculate the size of angle $B$.
In a triangle, the following data are known: $a = 12$ m, $b = 8$ m and angle $A = 120^\circ$. Calculate the size of angle $B$.
Seán has to calculate the length of side $a$ of a triangle for which he knows $b = 5\text{ cm}$, $c = 8\text{ cm}$, and the included angle $A = 120^\circ$. In his…
Seán has to calculate the length of side $a$ of a triangle for which he knows $b = 5\text{ cm}$, $c = 8\text{ cm}$, and the included angle $A = 120^\circ$. In his notebook, he writes the following resolution:
$$a^2 = 5^2 + 8^2 - 2 \times 5 \times 8 \times \cos(120^\circ)$$ $$a^2 = 25 + 64 - 80 \times (-0.5)$$ $$a^2 = 89 - 40 = 49 \implies a = 7\text{ cm}$$1) Point out the calculation error Seán made when working with the signs.
2) Without having to redo the calculations yet, explain why a result of $a = 7\text{ cm}$ is geometrically impossible in this triangle (relate the magnitude of angle $A$ to the relative measures of the sides).
3) Correct the procedure and calculate the correct value of side $a$ rounded to two decimal places.
Solving the complete triangle
"Solving" an oblique triangle means the same as in a right-angled one: calculating all three sides and all three angles, combining the two rules in whichever order is convenient.
When the given angle $A$ is acute and the side opposite it is shorter than the other given side ($a
In the design of an articulated robotic arm, the first link measures $40$ cm and the second measures $35$ cm. If the angle formed by the first link with the horizontal base is $50^\circ$ and the end of the second link touches the horizontal base exactly, calculate the two possible distances from the origin of the arm to the contact point on the base. In a triangle $ABC$, the angles $A$ and $B$ measure $30^\circ$ and $105^\circ$ respectively. If the side opposite angle $A$ has a length of $a = 8$ cm, find the length of side $c$ (opposite angle $C$). In an isosceles triangle, the non-equal side (the base) measures $12$ m and the equal angles measure $75^\circ$ each. Determine the length of the equal sides using the sine rule. In a triangle $ABC$, side $a$ measures $10$ cm and the opposite angle $A$ measures $45^\circ$. Calculate the radius $R$ of the circumscribed circle of the triangle.In the design of an articulated robotic arm, the first link measures $40$ cm and the second measures $35$ cm. If the angle formed by the first link with the horizontal…
In a triangle $ABC$, the angles $A$ and $B$ measure $30^\circ$ and $105^\circ$ respectively. If the side opposite angle $A$ has a length of $a = 8$ cm, find the length…
In an isosceles triangle, the non-equal side (the base) measures $12$ m and the equal angles measure $75^\circ$ each. Determine the length of the equal sides using the…
In a triangle $ABC$, side $a$ measures $10$ cm and the opposite angle $A$ measures $45^\circ$. Calculate the radius $R$ of the circumscribed circle of the triangle.
Area of a triangle without its height
If you know two sides and the angle between them, you do not need to calculate any height: there is a direct formula for the area.
Find the area of a triangle $ABC$ for which the sides $b = 14$ cm, $c = 20$ cm and the angle between them, $A = 60^\circ$, are known.
Find the area of a triangle $ABC$ for which the sides $b = 14$ cm, $c = 20$ cm and the angle between them, $A = 60^\circ$, are known.
The adjacent sides of a parallelogram measure $6$ cm and $10$ cm, forming an obtuse angle of $120^\circ$. Calculate the length of the longer diagonal of the…
The adjacent sides of a parallelogram measure $6$ cm and $10$ cm, forming an obtuse angle of $120^\circ$. Calculate the length of the longer diagonal of the parallelogram.
In a triangle, we know sides $a = 6\text{ cm}$ and $b = 6\text{ cm}$, and the included angle between them $C = 62^\circ$. 1) Estimate: before performing any numerical…
In a triangle, we know sides $a = 6\text{ cm}$ and $b = 6\text{ cm}$, and the included angle between them $C = 62^\circ$.
1) Estimate: before performing any numerical calculations with formulas, reason whether the third side $c$ will be slightly greater than, less than, or equal to $6\text{ cm}$. To justify this, imagine what the figure would look like if the angle were exactly $60^\circ$ (equilateral triangle) and analyze what happens when opening the angle two degrees more.
2) Check: calculate the exact length of side $c$ using the law of cosines (round to two decimal places).
3) Assess whether the obtained result confirms your qualitative estimate and calculate the absolute error you would have made if you had directly approximated $c \approx 6\text{ cm}$.
Heron's formula and the inscribed circle
If you know all three sides, let $s=\frac{a+b+c}{2}$ (the semi-perimeter). Then $S=\sqrt{s(s-a)(s-b)(s-c)}$. The inscribed circle (touching all three sides) has radius $r=\dfrac{S}{s}$, because the triangle splits into three triangles of height $r$ on the sides $a$, $b$, $c$, so $S=\frac12 r(a+b+c)=rs$. The circumscribed radius comes from $R=\dfrac{abc}{4S}$.
In a triangle $ABC$ with sides $a=13$ cm, $b=14$ cm and $c=15$ cm, the circumscribed circle of radius $R$ and the inscribed circle of radius $r$ are drawn. Calculate the…
In a triangle $ABC$ with sides $a=13$ cm, $b=14$ cm and $c=15$ cm, the circumscribed circle of radius $R$ and the inscribed circle of radius $r$ are drawn. Calculate the area of the triangle using Heron's formula and, from it, obtain the exact values of $R$ and $r$.
Triangulation in engineering and navigation
Solving oblique triangles is the basis of triangulation: measuring angles and distances from reference points to fix the position of a third, inaccessible point.
Back to the start: the buses leave the depot on routes 75° apart and cover 12 km and 16 km, so the distance between them is the third side of a triangle with two known sides and the included angle. The cosine rule solves it, exactly as with the two ships: $d^2=12^2+16^2-2\cdot12\cdot16\cdot\cos75^\circ\approx400-99.4\approx300.6$, so $d\approx17.3$ km.
Two ships leave a port simultaneously along straight routes that form an angle of $75^\circ$ with each other. The first ship sails at a constant speed of $15$ knots and…
Two ships leave a port simultaneously along straight routes that form an angle of $75^\circ$ with each other. The first ship sails at a constant speed of $15$ knots and the second at $20$ knots. Calculate the distance in nautical miles separating the two ships after $2$ hours of navigation.
To measure the height $h$ of a wind turbine, we stand at a point $A$ on the ground and measure the angle of elevation to the top point, obtaining $40^\circ$. We move…
To measure the height $h$ of a wind turbine, we stand at a point $A$ on the ground and measure the angle of elevation to the top point, obtaining $40^\circ$. We move $30$ metres in a straight line towards the base of the wind turbine to reach a point $B$, from which the new angle of elevation is $55^\circ$. Calculate the height of the wind turbine.
A surveillance drone is at a point $C$ observing a straight stretch of road bounded by two cities $A$ and $B$ that are $5$ km apart. The angles of depression from the…
A surveillance drone is at a point $C$ observing a straight stretch of road bounded by two cities $A$ and $B$ that are $5$ km apart. The angles of depression from the drone to the cities are such that in the triangle formed, the angle at $A$ is $35^\circ$ and the angle at $B$ is $62^\circ$. Calculate the altitude at which the drone is flying above the road.
Quadrilaterals and polygons: dividing into triangles
Any quadrilateral or polygon can be solved by dividing it into triangles with a diagonal, and applying the sine and cosine rules to each one.
A piece of farmland has the shape of a quadrilateral $ABCD$. The lengths of its sides are $AB = 50$ m, $BC = 40$ m, $CD = 30$ m and $DA = 60$ m. Knowing that the angle…
A piece of farmland has the shape of a quadrilateral $ABCD$. The lengths of its sides are $AB = 50$ m, $BC = 40$ m, $CD = 30$ m and $DA = 60$ m. Knowing that the angle at vertex $B$ is $70^\circ$, calculate the total area of the land by dividing it into two triangles.
A regular heptagon (a polygon with $7$ equal sides) is inscribed in a circle of radius $10$ cm. Using the cosine rule and/or sine rule in the central triangles,…
A regular heptagon (a polygon with $7$ equal sides) is inscribed in a circle of radius $10$ cm. Using the cosine rule and/or sine rule in the central triangles, calculate the perimeter and area of the heptagon.
The three sides of a triangle are known: $a = 4\text{ cm}$, $b = 7\text{ cm}$, and $c = 10\text{ cm}$. To solve it completely, you must find its three angles. Two…
The three sides of a triangle are known: $a = 4\text{ cm}$, $b = 7\text{ cm}$, and $c = 10\text{ cm}$. To solve it completely, you must find its three angles. Two classmates propose different paths to calculate the second angle after having obtained the first:
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Aoife: "First, I calculate the smallest angle, $A$, with the law of cosines. Then, to make the calculations faster and simpler, I calculate the largest angle, $C$, using the law of sines ($\frac{a}{\sin A} = \frac{c}{\sin C}$), and finally $B = 180^\circ - A - C$".
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Seán: "It is dangerous to calculate $C$ using the law of sines. It is much safer to calculate $C$ directly with the law of cosines, or use the law of sines to calculate $B$, but never for $C$".
1) Mathematically justify Seán's objection: what limitation does the $\arcsin$ function on the calculator have when looking for angles that could be obtuse?
2) Check the flaw in Aoife's method: calculate angle $A$ using the law of cosines and show what incorrect value for $C$ the law of sines would yield if applied directly.
3) Apply the safe path proposed by Seán to calculate the three angles of the triangle accurately.
Projection formula and angle bisectors
Draw the altitude from $A$ to side $a$. It splits $a$ into $c\cos B$ and $b\cos C$, so $a=b\cos C+c\cos B$ (the projection formula).
To get the compound angle formula: by the sine rule $a=2R\sin A$, $b=2R\sin B$, $c=2R\sin C$. Substitute: $\sin A=\sin B\cos C+\cos B\sin C$. Since $A=180^\circ-(B+C)$, $\sin A=\sin(B+C)$.
For an angle bisector of $C$ meeting $AB$ at $D$, with $CD=x$ and $C=2\gamma$: the areas add up, $\frac12ab\sin2\gamma=\frac12 bx\sin\gamma+\frac12 ax\sin\gamma$. Since $\sin2\gamma=2\sin\gamma\cos\gamma$, we get $x=\dfrac{2ab\cos\gamma}{a+b}$.
Prove geometrically or using trigonometric rules (sine and cosine rules) that in any triangle $ABC$ the projection formula holds: $a = b \cos C + c \cos B$. Next, use…
Prove geometrically or using trigonometric rules (sine and cosine rules) that in any triangle $ABC$ the projection formula holds: $a = b \cos C + c \cos B$. Next, use this identity together with the sine rule to prove the sine compound angle formula: $\sin(B+C) = \sin B \cos C + \cos B \sin C$.
Given the triangle $ABC$ with sides $a = 7$ cm, $b = 5$ cm and $c = 8$ cm, the internal angle bisector of angle $C$ is drawn, intersecting side $AB$ at point $D$.…
Given the triangle $ABC$ with sides $a = 7$ cm, $b = 5$ cm and $c = 8$ cm, the internal angle bisector of angle $C$ is drawn, intersecting side $AB$ at point $D$. Applying the cosine rule to find angle $C$ and using the properties of areas or the sine rule, calculate the exact length of the angle bisector segment $CD$ (denoted as $x$).
You can do it on paper and send a photo with your answer.
Problems in 3D
In 3D, pick a flat cross-section containing the unknown, draw it as a plane triangle, and apply the sine or cosine rule there. Mark which lengths you already know from the solid.
A regular square-based pyramid has a base side length of $100$ m. Knowing that the dihedral angle formed by two adjacent triangular faces is $105^\circ$, set up the…
A regular square-based pyramid has a base side length of $100$ m. Knowing that the dihedral angle formed by two adjacent triangular faces is $105^\circ$, set up the geometry of the appropriate cross-section and calculate the total height of the pyramid.