In this page, we will look at the Sine Rule, the Cosine Rule and the formula for the Area of a Triangle. The formula for these three are fairly straight forward to use, but don't take this topic too lightly, sometimes the exam questions can be quite challenging. Questions on this topic can even come up on the non-calculator paper when we use the angles 30°, 45°, 60°, ...
On this page, you should learn about
- the sine rule: \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\)
- the cosine rule: \(c^2=a^2+b^2-2ab\cos C\)
- the area of a triangle =\({ 1\over 2}ab\sin C\)
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Here is a quiz that practises the skills from this page
START QUIZ!
Using the following diagram, complete the following equation
\(\frac{5}{sinA°}=\frac{b}{sin57°}\)
A =
b =
Using the Sine Rule, a is opposite A and b is opposite B
\(\frac{a}{sinA}=\frac{b}{sinB}\)
Using the following diagram, complete the following equation
\(\frac{5}{sinA°}=\frac{8}{sinB°}\)
A =
B =
The side with length 8 is opposite the unmarked angle. We calculate 180-30-98 = 52
In the triangle ABC, BC = 7cm , AC = 7.5 cm and \(\hat { BAC } \)= 66°
If we use the Sine Rule to calculate \(\hat { BAC } \) , one solution is 78°.
Find another possible solution
\(\hat { BAC } \) = °
This is the ambiguous case for the Sine Rule since sin78° = sin(180°-78°)
Using the following diagram, complete the following equation
² = 5² + ² - 2 x x 8 x cos 105°
Using the Cosine Rule
c² = a² + b² - 2abcosC
The side c is opposite the angle C
Using the following diagram, complete the following equation
cos A° = \(\frac{3^2+b^2-6.2^2}{2\times3\times c}\)
A =
b =
c =
cos 124° = \(\frac{3^2+4^2-6.2^2}{2\times3\times 4}\)
In the triangle ABC, AC = 4cm , BC = 6cm and angle \(\hat { B } \) = 30°
a) Work out sin \(\hat { A } \)
There are two possible answers for angle A, A1 and A2 where A1 < A2
b) Find A1 and A2 to 3 significant figures
a) sin \(\hat { A } \) =
b) A1 = ° , A2 = °
Using the Sine Rule
\(\frac{4}{sin30°}=\frac{6}{sinA}\)
\(sinA=\frac{6\times sin30°}{4}\)
\(sinA=\frac{6\times 0.5}{4}\)
sinA = 0.75
b) Ambiguous Case
A1 = arcsin(0.75)=48.6°
A2 = 180°- 48.6°=131.4°
Rounded to 3 s.f. A2 = 131°
In the triangle ABC, AB = 8cm, AC = \(2 \sqrt{3} \) cm and \(\hat { BAC } \) = 30°.
If BC = \(a \sqrt{b} \) , work out a and b
a =
b =
Using Cosine Rule
BC² = 8² + (\(2 \sqrt{3} \))² - 2x \(2 \sqrt{3} \) x 8 x cos 30°
BC² = 64 + 4x3 - 2 x \(2 \sqrt{3} \) x 8 x \(\frac{\sqrt{3}}{2}\)
BC² = 64 + 12 - 2 x 3 x 8
BC² = 28
BC = \( \sqrt{28}\)
BC = \( 2\sqrt{7}\)
In the triangle ABC, AC = 12 cm , \(\hat{ABC}\) = 45° and \(\hat{ACB}\) = 75°
If BC = \(a \sqrt{b} \) , work out a and b
a =
b =
Using the Sine Rule
\(\frac{BC}{sin60°}=\frac{12}{sin45°}\)
\(BC=\frac{12\times sin60°}{sin45°}\)
\(BC=\frac{12\times\frac { \sqrt { 3 } }{ 2 } }{\frac { 1 }{ \sqrt { 2 } } }\)
\(BC=\frac { 6\sqrt { 3 } }{ \frac { 1 }{ \sqrt { 2 } } } \)
\(BC = 6\sqrt { 3 } \times \frac { \sqrt { 2 } }{ 1 } \)
\(BC = 6\sqrt { 6 }\)
In the triangle ABC, AB = \(3{\sqrt{2}}\) cm , AC = 8 cm and \(\hat{BAC}\) = 135°
Find the area of the triangle ABC
Area = cm²
Area = \(\frac { 1 }{ 2 } \times 8\times 3\sqrt { 2 } \times sin135°\)
= \(\frac { 1 }{ 2 } \times 8\times 3\sqrt { 2 } \times \frac { 1 }{ \sqrt { 2 } } \)
= 12
In the triangle ABC, the area = 52 cm², AB = 12cm and AC = 13 cm.
There are two possible values for angle A, A1 and A2 where A1 < A2
Find A1 and A2 to 3 significant figures
A1 = °
A2 = °
Area = 0.5 x 12 x 13 x sinA
52 = 78 x sin A
sinA = \(\frac{52}{78}\)
A = \(arcsin(\frac{52}{78})\) = 41.8°
A2 = 180° - 41.8° = 138.2°
To 3 s.f. A2 = 138°
The following diagram shows a quadrilateral ABCD.
AB = 7cm , AD = 5cm ∠DAB=120° , ∠DBC=45° , ∠BCD=60°
BD = \(\sqrt{a}\)
CD = \(\sqrt{b}\)
\(a,b \in \mathbb{Z}\)
Find a and b
Hint
Full Solution
The following diagram shows a quadrilateral ABCD.
AD = x – 1 , BD = x + 1 , DC = 2x and \(\angle CDA\) = 120°
The sum of the area of triangle ADC and triangle BDC is \(4 \sqrt{3}\)
Find x
Hint
Full Solution
How much of Sine and Cosine Rule have you understood?
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