Solution of Triangles Worked Examples (KBAT)

Q: How do I know when to look for a second triangle?

The ambiguous case only arises with two sides and a non-included angle (the SSA arrangement). If the side opposite the given angle is shorter than the other given side, test both the acute angle and its obtuse partner, keeping any that leave the angle sum below 180∘180^{\circ}180∘. With an included angle, or when the opposite side is the longer one, the triangle is unique.

Q: Why does the cosine rule sometimes give two answers here?

Written as a quadratic in the unknown side, c2−2bcos⁡A c+(b2−a2)=0c^{2}-2b\cos A\,c+(b^{2}-a^{2})=0c2−2bcosAc+(b2−a2)=0, the cosine rule can have two positive roots. Each root is the third side of one of the two triangles that fit the data, which is the same ambiguity the sine rule shows through an acute and an obtuse angle.

Q: Is the perpendicular distance from a point to a line always a height?

Yes, when the point is a vertex and the line contains the opposite side: that perpendicular is the triangle's height for that base. You can find it from Area=12×base×height\text{Area}=\tfrac{1}{2}\times\text{base}\times\text{height}Area=21×base×height once you have the area, or directly as a known side times the sine of the angle it makes with the base.

Q: What tells me an angle is obtuse before I press the calculator?

A negative cosine. When the rearranged cosine rule gives cos⁡θ<0\cos\theta<0cosθ<0, the angle lies between 90∘90^{\circ}90∘ and 180∘180^{\circ}180∘. Its sine stays positive, so the area formula is unaffected, and the side opposite that angle will be the longest in the triangle.

What these examples cover

These hard Solution of Triangles examples ask you to hold two ideas at once. The first is the ambiguous case: given two sides and a non-included angle, the sine rule can allow two different triangles, and you have to test whether both really fit.

The second continues an area calculation into a perpendicular height, joining the area formula to the idea of a base and its height. The third is a surveyed plot cut by a diagonal into two triangles, where one triangle feeds a shared side into the next and an obtuse angle appears.

The numbers stay clean enough to follow, but the reasoning is where the marks live. Use the set the honest way, cover the solution, attempt each question in full on paper, and only then check line by line.

Where your answer differs, decide first whether the gap is arithmetic or a genuine choice, such as an acute-versus-obtuse angle. A careful labelled sketch is worth more here than in any easier set.

Worked examples

Work through all three. Each rewards a plan made before the first line of algebra: decide which rule starts the question, what the intermediate result will be, and how you will check it independently at the end.

Q1[6 marks]

In triangle $ABC$ , $a = 6$ cm, $b = 8$ cm and $\angle A = 40^{\circ}$ . Show that two different triangles fit this data, and find $\angle B$ , $\angle C$ and side $c$ for each.

Show worked solution

You are given two sides and a non-included angle (the angle $A$ does not sit between $a$ and $b$ ). This is the ambiguous case, so begin with the sine rule for $\angle B$ and stay alert for a second solution:

\sin B=\dfrac{b\sin A}{a}=\dfrac{8\sin 40^{\circ}}{6}=\dfrac{8\times0.6428}{6}=0.8571

Two angles between $0^{\circ}$ and $180^{\circ}$ share this sine, an acute one and its obtuse partner $180^{\circ}$ minus it:

B_{1}=\sin^{-1}(0.8571)=58.99^{\circ}\qquad B_{2}=180^{\circ}-58.99^{\circ}=121.01^{\circ}

Both survive the angle-sum test, because in each case $A+B<180^{\circ}$ : $40^{\circ}+58.99^{\circ}=98.99^{\circ}$ and $40^{\circ}+121.01^{\circ}=161.01^{\circ}$ . So two triangles genuinely fit.

Take each in turn.

Triangle 1 (with $B_{1}=58.99^{\circ}$ ): the third angle and the last side are

C_{1}=180^{\circ}-40^{\circ}-58.99^{\circ}=81.01^{\circ},\quad c_{1}=\dfrac{a\sin C_{1}}{\sin A}=\dfrac{6\sin 81.01^{\circ}}{\sin 40^{\circ}}=9.22

Triangle 2 (with $B_{2}=121.01^{\circ}$ ):

C_{2}=180^{\circ}-40^{\circ}-121.01^{\circ}=18.99^{\circ},\quad c_{2}=\dfrac{6\sin 18.99^{\circ}}{\sin 40^{\circ}}=3.04

Answer

Triangle 1: $\angle B=58.99^{\circ}$ , $\angle C=81.01^{\circ}$ , $c=9.22$ cm. Triangle 2: $\angle B=121.01^{\circ}$ , $\angle C=18.99^{\circ}$ , $c=3.04$ cm.

A neat check treats the cosine rule as a quadratic in $c$ : $c^{2}-2b\cos A\,c+(b^{2}-a^{2})=0$ becomes $c^{2}-12.2567c+28=0$ , whose two roots $9.22$ and $3.04$ are exactly the two values of $c$ .

When is the ambiguous case really ambiguous?

Only when you are given two sides and a non-included angle, and the side opposite the given angle is the shorter of the two. Here $a=6$ faces the $40^{\circ}$ angle and is shorter than $b=8$ , which opens the door to two triangles.

If the side opposite the given angle were the longer one, only the acute solution would survive.

Q2[7 marks]

In triangle $ABC$ , $AB = 8$ cm, $AC = 15$ cm and $\angle A = 60^{\circ}$ . Find (a) the length of $BC$ , (b) the area of the triangle, and (c) the perpendicular distance from $B$ to $AC$ .

Show worked solution

(a) Two sides with the included angle $A$ point to the cosine rule. With $BC$ opposite $A$ :

BC^{2}=AB^{2}+AC^{2}-2(AB)(AC)\cos A=8^{2}+15^{2}-2(8)(15)\cos 60^{\circ}

BC^{2}=64+225-240(0.5)=289-120=169\;\Rightarrow\; BC=\sqrt{169}=13

(b) The area uses the two sides that enclose $A$ , namely $AB$ and $AC$ , with $\sin 60^{\circ}=0.8660$ :

\text{Area}=\tfrac{1}{2}(AB)(AC)\sin A=\tfrac{1}{2}(8)(15)(0.8660)=60\times0.8660=51.96

(c) The perpendicular distance from $B$ to $AC$ is the height of the triangle when $AC$ is taken as the base. Since the area equals $\tfrac{1}{2}\times\text{base}\times\text{height}$ , solve for the height $h$ :

51.96=\tfrac{1}{2}(15)(h)\;\Rightarrow\; h=\dfrac{2\times51.96}{15}=\dfrac{103.92}{15}=6.93

Answer

$BC = 13$ cm, the area is $51.96$ cm $^{2}$ , and the perpendicular distance from $B$ to $AC$ is $6.93$ cm (2 d.p.). Check the height directly: dropping a perpendicular from $B$ onto $AC$ makes a right triangle with hypotenuse $AB=8$ and angle $A=60^{\circ}$ , so the height is $AB\sin A=8\sin 60^{\circ}=6.93$ , which agrees.

Q3[7 marks]

A surveyor records a quadrilateral plot $ABCD$ . In triangle $ABC$ , $AB = 3$ m, $BC = 5$ m and $\angle ABC = 120^{\circ}$ .

In triangle $ACD$ , the sides are $AC$ , $CD = 4$ m and $AD = 9$ m. Find (a) the length of the diagonal $AC$ , and (b) the total area of the plot $ABCD$ .

Show worked solution

(a) The diagonal $AC$ is shared by both triangles, so find it first from triangle $ABC$ . It is opposite the given angle $\angle ABC$ , so use the cosine rule with that angle.

As $120^{\circ}$ is obtuse, $\cos 120^{\circ}=-0.5$ :

AC^{2}=AB^{2}+BC^{2}-2(AB)(BC)\cos 120^{\circ}=3^{2}+5^{2}-2(3)(5)(-0.5)

AC^{2}=9+25+15=49\;\Rightarrow\; AC=\sqrt{49}=7

(b) Find each triangle's area and add them. Triangle $ABC$ has the included angle at $B$ between $AB$ and $BC$ :

\text{Area}_{ABC}=\tfrac{1}{2}(AB)(BC)\sin 120^{\circ}=\tfrac{1}{2}(3)(5)\left(\tfrac{\sqrt{3}}{2}\right)=\tfrac{15\sqrt{3}}{4}=6.4952

Triangle $ACD$ has three known sides $AC=7$ , $CD=4$ and $AD=9$ , so recover an angle with the cosine rule. Angle $ACD$ , at $C$ between $CA$ and $CD$ , faces $AD=9$ :

\cos(\angle ACD)=\dfrac{AC^{2}+CD^{2}-AD^{2}}{2(AC)(CD)}=\dfrac{49+16-81}{2(7)(4)}=\dfrac{-16}{56}=-\dfrac{2}{7}

The negative cosine shows the angle is obtuse: $\angle ACD=\cos^{-1}\!\left(-\tfrac{2}{7}\right)=106.60^{\circ}$ . Take its sine from $\sin^{2}+\cos^{2}=1$ , which avoids rounding the angle: $\sin(\angle ACD)=\sqrt{1-\left(\tfrac{2}{7}\right)^{2}}=\tfrac{\sqrt{45}}{7}\approx0.9583$ .

Its area is

\text{Area}_{ACD}=\tfrac{1}{2}(AC)(CD)\sin(\angle ACD)=\tfrac{1}{2}(7)(4)\left(\tfrac{\sqrt{45}}{7}\right)=2\sqrt{45}=13.4164

Add the two areas, keeping full precision until the final line:

\text{Total}=6.4952+13.4164=19.9116

Answer

The diagonal $AC = 7$ m and the total area of the plot is $19.91$ m $^{2}$ (2 d.p.). Keep the unrounded areas until the end, adding the rounded parts $6.50+13.42$ gives $19.92$ , a small rounding artefact rather than a better answer.

Q4[6 marks]

A ship leaves port $P$ and sails on a bearing of $030^{\circ}$ for $7$ km to reach point $A$ . At the same time, a second ship leaves $P$ and sails on a bearing of $130^{\circ}$ for $9$ km to reach point $B$ .

Find (a) the distance $AB$ , and (b) the bearing of $B$ from $A$ .

Show worked solution

Both bearings are measured from north at $P$ , so the angle between the two ships' paths is their difference: $\angle APB=130^{\circ}-030^{\circ}=100^{\circ}$ . With $PA=7$ and $PB=9$ enclosing this angle, use the cosine rule for (a).

AB^{2}=PA^{2}+PB^{2}-2(PA)(PB)\cos100^{\circ}=7^{2}+9^{2}-2(7)(9)(-0.1736)

AB^{2}=49+81+21.88=151.88\;\Rightarrow\;AB=\sqrt{151.88}=12.32\text{ km}

(b) Find $\angle PAB$ first with the sine rule, then convert it into a bearing:

\sin(\angle PAB)=\dfrac{PB\sin(\angle APB)}{AB}=\dfrac{9\sin100^{\circ}}{12.32}=\dfrac{9\times0.9848}{12.32}=0.7192\;\Rightarrow\;\angle PAB=45.99^{\circ}

The bearing of $P$ from $A$ is the reverse of $A$ 's bearing from $P$ , $030^{\circ}+180^{\circ}=210^{\circ}$ . A sketch shows $B$ lies on the side where turning by $\angle PAB$ reduces this bearing, so:

\text{Bearing of }B\text{ from }A=210^{\circ}-45.99^{\circ}=164.0^{\circ}

Answer

$AB=12.32$ km and the bearing of $B$ from $A$ is $164.0^{\circ}$ . Check: the third angle $\angle PBA=180^{\circ}-100^{\circ}-45.99^{\circ}=34.01^{\circ}$ gives $PA/\sin(\angle PBA)=7/\sin34.01^{\circ}\approx12.51$ , matching $AB/\sin100^{\circ}\approx12.51$ .

Q5[6 marks]

A vertical tower stands with its foot at point $Q$ on horizontal ground. From a point $A$ on the ground, $AQ=15$ m and the angle of elevation of the top of the tower, $T$ , from $A$ is $28^{\circ}$ .

A second point $B$ also lies on the ground, with $AB=20$ m and $\angle QAB=65^{\circ}$ . Find (a) the distance $BQ$ , (b) the height of the tower, and (c) the angle of elevation of $T$ from $B$ .

Show worked solution

(a) Points $A$ , $B$ and $Q$ all lie on the horizontal ground, so triangle $ABQ$ is an ordinary triangle. With $AQ$ and $AB$ enclosing the given angle, use the cosine rule:

BQ^{2}=AQ^{2}+AB^{2}-2(AQ)(AB)\cos65^{\circ}=15^{2}+20^{2}-2(15)(20)(0.4226)

BQ^{2}=225+400-253.56=371.44\;\Rightarrow\;BQ=\sqrt{371.44}=19.27\text{ m}

(b) $TQ$ is vertical, so triangle $AQT$ is right-angled at $Q$ , and the angle of elevation from $A$ relates the height directly to $AQ$ :

TQ=AQ\tan28^{\circ}=15\times0.5317=7.98\text{ m}

(c) The same reasoning applies in right-angled triangle $BQT$ , using the now-known height and the distance $BQ$ found in (a):

\tan(\angle TBQ)=\dfrac{TQ}{BQ}=\dfrac{7.98}{19.27}=0.4141\;\Rightarrow\;\angle TBQ=22.49^{\circ}

Answer

$BQ=19.27$ m, the tower is $7.98$ m tall, and the angle of elevation from $B$ is $22.49^{\circ}$ . Sense check: $B$ is farther from the foot of the tower than $A$ ( $19.27>15$ ), so it should see a smaller angle of elevation, $22.49^{\circ}<28^{\circ}$ , as expected.

Q6[5 marks]

Three lamp posts stand at points $A$ , $B$ and $C$ , to be linked by a single circular path passing through all three (the circle that circumscribes triangle $ABC$ ). In triangle $ABC$ , $\angle B=72^{\circ}$ , $\angle C=48^{\circ}$ and $AC=10$ m.

Find (a) the lengths of $AB$ and $BC$ , and (b) the radius of the circular path.

Show worked solution

The three angles sum to $180^{\circ}$ , so $\angle A=180^{\circ}-72^{\circ}-48^{\circ}=60^{\circ}$ . Side $AC$ is opposite $\angle B$ , so the sine rule's constant ratio (twice the circumradius) is fixed by the pair you already know:

2R=\dfrac{AC}{\sin B}=\dfrac{10}{\sin72^{\circ}}=\dfrac{10}{0.9511}=10.51\text{ m}

(a) Every side equals $2R$ times the sine of its opposite angle, so $BC$ (opposite $A$ ) and $AB$ (opposite $C$ ) follow at once:

BC=2R\sin A=10.51\times\sin60^{\circ}=10.51\times0.8660=9.11\text{ m}

AB=2R\sin C=10.51\times\sin48^{\circ}=10.51\times0.7431=7.81\text{ m}

(b) The radius of the circular path is half this constant:

R=\dfrac{10.51}{2}=5.26\text{ m}

Answer

$AB\approx7.81$ m, $BC\approx9.11$ m, and the circular path has radius $R\approx5.26$ m. Check: computing the same ratio from $BC$ gives $BC/\sin A=9.11/\sin60^{\circ}\approx10.52$ , matching $2R\approx10.51$ from the original ratio.

Q7[6 marks]

In triangle $ABD$ , $AB=10$ cm, $\angle ABD=50^{\circ}$ and $\angle ADB=100^{\circ}$ , where $D$ lies on the straight line $BC$ . Given $DC=6$ cm, find (a) the length of $AD$ , (b) the length of $AC$ , and (c) $\angle BAC$ of triangle $ABC$ .

Show worked solution

(a) The angles of triangle $ABD$ sum to $180^{\circ}$ , so $\angle BAD=180^{\circ}-50^{\circ}-100^{\circ}=30^{\circ}$ . Use the sine rule to find $AD$ :

AD=\dfrac{AB\sin(\angle ABD)}{\sin(\angle ADB)}=\dfrac{10\sin50^{\circ}}{\sin100^{\circ}}=\dfrac{10\times0.7660}{0.9848}=7.78\text{ cm}

(b) Because $D$ sits on the straight line $BC$ , $\angle ADC$ is the supplement of $\angle ADB$ : $\angle ADC=180^{\circ}-100^{\circ}=80^{\circ}$ . Triangle $ADC$ now has two sides and the included angle, so use the cosine rule:

AC^{2}=AD^{2}+DC^{2}-2(AD)(DC)\cos80^{\circ}=7.78^{2}+6^{2}-2(7.78)(6)(0.1736)

AC^{2}=60.53+36-16.21=80.32\;\Rightarrow\;AC=\sqrt{80.32}=8.96\text{ cm}

\sin(\angle DAC)=\dfrac{DC\sin(\angle ADC)}{AC}=\dfrac{6\sin80^{\circ}}{8.96}=\dfrac{6\times0.9848}{8.96}=0.6594\;\Rightarrow\;\angle DAC=41.25^{\circ}

\angle BAC=\angle BAD+\angle DAC=30^{\circ}+41.25^{\circ}=71.25^{\circ}

Answer

$AD\approx7.78$ cm, $AC\approx8.96$ cm and $\angle BAC\approx71.25^{\circ}$ . Check: the three angles of triangle $ABC$ are $50^{\circ}$ at $B$ , $180^{\circ}-80^{\circ}-41.25^{\circ}=58.75^{\circ}$ at $C$ , and $71.25^{\circ}$ at $A$ , these sum to exactly $180^{\circ}$ .

Q8[5 marks]

A triangular garden plot has sides $7$ m, $9$ m and $13$ m. (a) Without finding its exact size, determine whether the largest angle of the plot is acute or obtuse, giving a reason.

(b) Find the size of the largest angle. (c) Find the area of the plot.

Show worked solution

(a) The largest angle is always opposite the longest side, so it is the angle $C$ opposite the $13$ m side. Testing the sign of the cosine rule's numerator settles the type without solving for the angle:

a^{2}+b^{2}-c^{2}=7^{2}+9^{2}-13^{2}=49+81-169=-39

Since $a^{2}+b^{2}-c^{2}<0$ , the cosine of the largest angle is negative, so $C$ is obtuse.

(b) Divide by $2ab$ to get $\cos C$ , then take the inverse cosine:

\cos C=\dfrac{-39}{2(7)(9)}=\dfrac{-39}{126}=-0.3095\;\Rightarrow\;C=\cos^{-1}(-0.3095)=108.03^{\circ}

\text{Area}=\tfrac{1}{2}(7)(9)\sin108.03^{\circ}=31.5\times0.9509=29.95\text{ m}^{2}

Answer

The largest angle is obtuse, $C\approx108.03^{\circ}$ , and the area of the plot is $29.95$ m $^{2}$ . Check: Heron's formula with $s=\tfrac{7+9+13}{2}=14.5$ gives $\sqrt{14.5\times7.5\times5.5\times1.5}=\sqrt{897.19}\approx29.95$ , the same area.

The lesson across these three is that a hard question is rarely harder arithmetic, it is a decision made correctly. Ask whether the data can fit more than one triangle, whether an angle is acute or obtuse, and whether a length is a side or a height.

Make the decision explicit on the page, keep intermediate values unrounded, and finish with an independent check. That is how a demanding question becomes a steady sequence of ordinary steps.

Key method points

These three examples rehearse the judgement that hard Solution of Triangles questions test. Keep the following points in mind as you practise more.

The ambiguous case appears with two sides and a non-included angle; when the side opposite the given angle is the shorter one, test both the acute angle and its obtuse partner $180^{\circ}$ minus it.
A candidate second triangle is only valid if the angle sum stays under $180^{\circ}$ ; check $A+B<180^{\circ}$ before you keep it.
Treating the cosine rule as a quadratic in the unknown side, $c^{2}-2b\cos A\,c+(b^{2}-a^{2})=0$ , delivers both ambiguous-case lengths at once and cross-checks the sine-rule work.
A perpendicular distance from a vertex to the opposite side is a height: use $\text{Area}=\tfrac{1}{2}\times\text{base}\times\text{height}$ , or read it directly as (a side) $\times\sin(\text{angle})$ .
Split a quadrilateral along a diagonal, solve the triangle that carries the given angle first, then pass the shared side into the second triangle.
A negative cosine flags an obtuse angle; its sine is still positive, so the area formula $\tfrac{1}{2}ab\sin C$ works unchanged. Keep intermediate values unrounded, since analytic marking rewards a clear method even when the final digit is delicate.

How a teacher helps

The hardest questions here turn on a single decision, whether a second triangle exists, or whether a length is a side or a height, and that is exactly where a good teacher earns their keep. In a one-to-one lesson our teacher pauses at the decision point, asks you to argue both possibilities aloud, and only then lets the algebra run, so the judgement becomes yours rather than a memorised rule.

Because our teachers are experienced, you work with someone who has seen where these questions trip students and heads it off early. Lessons are taught in English, while SPM papers are set in both Malay and English, so the notation reads the same to you either way.

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Frequently asked questions

How do I know when to look for a second triangle?

The ambiguous case only arises with two sides and a non-included angle (the SSA arrangement). If the side opposite the given angle is shorter than the other given side, test both the acute angle and its obtuse partner, keeping any that leave the angle sum below $180^{\circ}$ .

With an included angle, or when the opposite side is the longer one, the triangle is unique.

Why does the cosine rule sometimes give two answers here?

Written as a quadratic in the unknown side, $c^{2}-2b\cos A\,c+(b^{2}-a^{2})=0$ , the cosine rule can have two positive roots. Each root is the third side of one of the two triangles that fit the data, which is the same ambiguity the sine rule shows through an acute and an obtuse angle.

Is the perpendicular distance from a point to a line always a height?

Yes, when the point is a vertex and the line contains the opposite side: that perpendicular is the triangle's height for that base. You can find it from $\text{Area}=\tfrac{1}{2}\times\text{base}\times\text{height}$ once you have the area, or directly as a known side times the sine of the angle it makes with the base.

What tells me an angle is obtuse before I press the calculator?

A negative cosine. When the rearranged cosine rule gives $\cos\theta<0$ , the angle lies between $90^{\circ}$ and $180^{\circ}$ .

Its sine stays positive, so the area formula is unaffected, and the side opposite that angle will be the longest in the triangle.

Source:SRC-DSKP-EN

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026

Solution of Triangles, Worked Examples (KBAT)

What these examples cover

Worked examples

Key method points

How a teacher helps

Frequently asked questions

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