Questions and Answers

1. Sine Rule IB Math Studies

IB Mathematical Studies – Trigonometry, Sine Rule

How can we find the side $a$ of a triangle ABC given that $\hat{A}=30^{\circ} , c=6 \ and\ \hat{C}=60^{\circ}$ ?

Solution

IB Maths SL – Trigonometry, Sine Rule

The Sine rule states

$\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}$

So, in this the side $a$ can be calculated as follows:

$\frac{a}{sin30^{\circ}}=\frac{6}{sin60^{\circ}}$

$a=\frac{1}{2} \frac{6}{\frac{\sqrt 3}{2}}=\frac{6}{ \sqrt 3}$

2. IB Math Studies Cosine Rule

IB Mathematical Studies – Trigonometry, Cosine Rule

How can we find the side $a$ of a triangle ABC given that $\hat{A}=60^{\circ} , c=4 \ and \ b=5$ ?

Solution

IB Mathematical Studies – Trigonometry, Cosine Rule

The Cosine rule states

$a^2=b^2+c^2-2(bc)cosA$

So, in this case the side $a$ can be calculated as follows:

$a^2=4^2+5^2-2(4(5))cos60^{\circ}$

$a^2=16+25-40 \frac{1}{2} =41-20=21$

$a=\sqrt{21}$

3. Area of a triangle Math Studies

IB Mathematical – Trigonometry, Area of a triangle

How can we find the area of a triangle ABC with sides $a=3 \ cm, b=1 \ cm$ and included angle $\hat{C}=30^{\circ}$ ?

Solution

IB Mathematical – Trigonometry, Area of a triangle

The area of a triangle is a half of the product of two sides and the sine of the included angle.

$A=\frac{1}{2}absinC$

Therefore, in this case, we have the following

$A=\frac{1}{2}(3)(1)sin30^{\circ} =\frac{3}{2} \frac{1}{2}=\frac{3}{4} \ cm^2$

IB Math Studies – Trigonometry