Given are two congruent triangles pqr and xyz. which of the following statements is incorrect?

In the module Further trigonometry (Year 10), we introduced and proved the sine rule, which is used to find sides and angles in non-right-angled triangles.

In the triangle \(ABC\), labelled as shown, we have

Clearly, we may also write this as

In general, one of the three angles may be obtuse. The formula still holds true, although the geometric proof is slightly different.

Exercise 4

1. Find two expressions for \(h\) in the diagram below, and hence deduce the sine rule.
   
   
   Detailed description of diagram

2. Repeat the method of part (a), using the following diagram, to show that the sine rule holds in obtuse-angled triangles.
   

The triangle \(ABC\) has \(AB = 9\text{ cm}\), \(\angle ABC = 76^\circ\) and \(\angle ACB = 58^\circ\).

Find, correct to two decimal places,

Solution

1. Applying the sine rule gives \[ \dfrac{AC}{\sin 76^\circ} = \dfrac{9}{\sin 58^\circ} \] and so \begin{align*} AC = \dfrac{9 \sin 76^\circ}{\sin 58^\circ}\\ \approx 10.30\text{ cm} \qquad \text{(to two decimal places).} \end{align*}
2. To find \(BC\), we first find the angle \(\angle CAB\) opposite it. \begin{align*} \angle CAB = 180^\circ - 58^\circ - 76^\circ\\ = 46^\circ. \end{align*} Thus, by the sine rule, \[ \dfrac{BC}{\sin 46^\circ} = \dfrac{9}{\sin58^\circ} \] and so \[ BC \approx 7.63\text{ cm} \qquad \text{(to two decimal places).} \]

The ambiguous case

In the module Congruence (Year 8), it was emphasised that, when applying the SAS congruence test, the angle in question has to be the angle included between the two sides. For example, the following diagram shows two non-congruent triangles \(ABC\) and \(ABC'\) having two pairs of matching sides and sharing a common (non-included) angle.

Suppose we are told that a triangle \(PQR\) has \(PQ = 9\), \(\angle PQR = 45^\circ\) and \(PR = 7\). Then the angle opposite \(PQ\) is not uniquely determined. There are two non-congruent triangles that satisfy the given data.

Applying the sine rule to the triangle, we have

Thus \(\theta \approx 65^\circ\), assuming that \(\theta\) is acute. But the supplementary angle is \(\theta' \approx 115^\circ\). The triangle \(PQR'\) also satisfies the given data. This situation is sometimes referred to as the ambiguous case.

Since the angle sum of a triangle is \(180^\circ\), in some circumstances only one of the two angles calculated is geometrically valid.

The cosine rule

We know from the SAS congruence test that a triangle is completely determined if we are given two sides and the included angle. However, if we know two sides and the included angle in a triangle, the sine rule does not help us determine the remaining side or the remaining angles.

The second important formula for general triangles is the cosine rule.

Suppose \(ABC\) is a triangle and that the angles \(A\) and \(C\) are acute. Drop a perpendicular from \(B\) to the line interval \(AC\) and mark the lengths as shown in the following diagram.

In the triangle \(ABD\), applying Pythagoras' theorem gives

Also, in the triangle \(BCD\), another application of Pythagoras' theorem gives

Substituting this expression for \(h^2\) into the first equation and expanding,

Finally, from triangle \(BCD\), we have \(x = a\,\cos C\) and so

This last formula is known as the cosine rule.

Notice that, if \(C = 90^\circ\), then since \(\cos C = 0\) we obtain Pythagoras' theorem, and so we can regard the cosine rule as Pythagoras' theorem with a correction term.

The cosine rule is also true when the angle \(C\) is obtuse. But note that, in this case, the final term in the formula will produce a positive number, because the cosine of an obtuse angle is negative. Some care must be taken in this instance.

By relabelling the sides and angles of the triangle, we can also write the cosine rule as \(a^2 = b^2 + c^2 - 2bc\,\cos A\) and \(b^2 = a^2 + c^2 - 2ac\,\cos B\).

Find the value of \(x\) to one decimal place.

Solution

Applying the cosine rule gives

so \(x \approx 12.3\) (to one decimal place).

Finding angles

If the three sides of a triangle are known, then the three angles are uniquely determined. (This is the SSS congruence test.) Again, the sine rule is of no help in finding the three angles, since it requires the knowledge of (at least) one angle, but we can use the cosine rule instead.

We can substitute the three side lengths \(a\), \(b\), \(c\) into the formula \(c^2 = a^2 + b^2 - 2ab\,\cos C\), where \(C\) is the angle opposite the side \(c\), and then rearrange to find \(\cos C\) and hence \(C\).

Alternatively, we can rearrange the formula to obtain

and then substitute. Students may choose to rearrange the cosine rule or to learn this further formula. Using this form of the cosine rule often reduces arithmetical errors.

Recall that, in any triangle \(ABC\) labelled as shown, if \(a < b\), then \(\text{angle } A < \text{angle } B\).

A triangle has side lengths 6 cm, 8 cm and 11 cm. Find the smallest angle in the triangle.

Solution

The smallest angle in the triangle is opposite the smallest side.

Applying the cosine rule:

So \(\theta \approx 32.2^\circ\) (correct to one decimal place).

The area of a triangle

We saw in the module Introductory trigonometry (Years 9--10) that, if we take any triangle with two given sides \(a\) and \(b\) about a given (acute) angle \(\theta\), then the area of the triangle is

This formula also holds when \(\theta\) is obtuse.

Exercise 5

A triangle has two sides of length 5 cm and 4 cm containing an angle \(\theta\). Its area is 5 cm\(^2\). Find the two possible (exact) values of \(\theta\) and draw the two triangles that satisfy the given information.

Exercise 6

Write down two different expressions for the area of a triangle \(ABC\) and derive the sine rule from them.

Next page - Content - Trigonometric identities

Two triangles are said to be congruent if one can be placed over the other so that they coincide (fit together). This means that congruent triangles are exact copies of each other and when fitted together the sides and angles which coincide, called corresponding sides and angles, are equal.

In Figure \(\PageIndex{1}\), \(\triangle ABC\) is congruent to \(\triangle DEF\). The symbol for congruence is \(\cong\) and we write \(\triangle ABC \cong \triangle DEF\). \(\angle A\) corresponds to \(\angle D\), \(\angle B\) corresponds to \(\angle E\), and \(\angle C\) corresponds to \(\angle F\). Side \(AB\) corresponds to \(DE, BC\) corresponds to \(EF\), and \(AC\) corresponds to \(DF\).

Figure \(\PageIndex{1}\): \(\triangle ABC\) is congruent to \(\triangle DEF\).

In this book the congruence statement \(\triangle ABC \cong \triangle DEF\) will always be written so that corresponding vertices appear in the same order, For the triangles in Figure \(\PageIndex{1}\), we might also write \(\triangle BAC \cong \triangle EDF\) or \(\triangle ACB \cong \triangle DFE\) but never for example \(\triangle ABC \cong \triangle EDF\) nor \(\triangle ACB \cong \triangle DEF\). (Be warned that not all textbooks follow this practice, Many authors wil write the letters without regard to the order. If that is the case then we cannot tell which parts correspond from the congruence statement)

Therefore we can always tell which parts correspond just from the congruence statement. For example, given that \(\triangle ABC \cong \triangle DEF\), side \(AB\) corresponds to side \(DE\) because each consists of the first two letters, \(AC\) corresponds to DF because each consists of the first and last letters, \(BC\) corresponds to \(EF\) because each consists of the last two letters.

If \(\triangle PQR \cong \triangle STR\)

1. list the corresponding angles and sides;
2. find \(x\) and \(y\).

Solution

(1)

\(\begin{array} {rcll} {\underline{\triangle PQR}} & \ & {\underline{\triangle STR}} & {} \\ {\angle P} & = & {\angle S} & {\text{(first letter of each triangle in congruence statement)}} \\ {\angle Q} & = & {\angle T} & {\text{(second letter)}} \\ {\angle PRQ} & = & {\angle SRT} & {\text{(third letter. We don't write "}\angle R = \angle R \text{" since}} \\ {} & & {} & {\text{each }\angle R \text{ is different)}} \\ {PQ} & = & {ST} & {\text{(first two letters)}} \\ {PR} & = & {SR} & {\text{(firsst and last letters)}} \\ {QR} & = & {TR} & {\text{(last two letters)}} \end{array}\)

(2)

\(x = PQ = ST = 6\).

\(y = PR = SR = 8\).

Answer (2): \(x = 6, y = 8\).

Assuming \(\triangle I \cong \triangle II\), write a congruence statement for \(\triangle I\) and \(\triangle II\):

Solution

\(\begin{array} {rcll} {\triangle I} & \ & {\triangle II} & {} \\ {\angle A} & = & {\angle B} & {(\text{both = } 60^{\circ})} \\ {\angle ACD} & = & {\angle BCD} & {(\text{both = } 30^{\circ})} \\ {\angle ADC} & = & {\angle BDC} & {(\text{both = } 90^{\circ})} \end{array}\)

Therefore

Answer: \(\triangle ACD \cong \triangle BCD\).

Assuming \(\triangle I \cong \triangle II\), write a congruence statement for \(\triangle I\) and \(\triangle II\):

Solution

The angles that are marked the same way are assumed to be equal.

\(\begin{array} {rcll} {\underline{\triangle I}} & \ & {\underline{\triangle II}} & {} \\ {\angle A} & = & {\angle B} & {(\text{both marked with one stroke})} \\ {\angle ACD} & = & {\angle BCD} & {(\text{both marked with two strokes})} \\ {\angle ADC} & = & {\angle BDC} & {(\text{both marked with three strokes})} \end{array}\)

The relationships are the same as in Example \(\PageIndex{2}\).

Answer: \(\triangle ACD \cong \triangle BCD\).

1 - 4. For each pair of congruent triangles

(1) list the corresponding sides and angles;

(2) find \(x\) and \(y\).

1. \(\triangle ABC \cong \triangle DEF\).

2. \(\triangle PQR \cong \triangle STU\).

3. \(\triangle ABC \cong \triangle CDA\).

4. \(\triangle ABC \cong \triangle EDC\).

5 - 10. Write a congruence statement for each of the following. Assume the triangles are congruent and that angles or sides marked in the same way are equal.

5.

6.

7.

8.

9.

10.