Geometry and Trigonometry - Sine Rule

In triangle $ABC$, angle $A = 40^{\circ}$, angle $B = 65^{\circ}$, and side $a = 12$ cm. Calculate the length of side $b$.

$\frac{b}{\sin 65^{\circ}} = \frac{12}{\sin 40^{\circ}}$ $b = \frac{12 \times \sin 65^{\circ}}{\sin 40^{\circ}}$ $b \approx \frac{12 \times 0.9063}{0.6428}$ $b \approx 16.9 \text{ cm (to 3 sig. fig.)}$

To find a missing side, we use the Sine Rule form with side lengths in the numerator. We substitute the known values for $a$, $A$, and $B$, then rearrange the equation to solve for $b$.

In triangle $PQR$, side $p = 8$ cm, side $q = 10$ cm, and angle $P = 35^{\circ}$. Find the size of angle $Q$ (assume $Q$ is acute).

$\frac{\sin Q}{10} = \frac{\sin 35^{\circ}}{8}$ $\sin Q = \frac{10 \times \sin 35^{\circ}}{8}$ $\sin Q \approx \frac{10 \times 0.5736}{8}$ $\sin Q \approx 0.7170$ $Q = \sin^{-1}(0.7170) \approx 45.8^{\circ}$

To find a missing angle, we use the Sine Rule form with sines in the numerator. After substituting the values and calculating $\sin Q$, we use the inverse sine function ($\sin^{-1}$) to find the angle.

A surveyor at point $A$ measures the angle of elevation to the top of a tower $T$ as $25^{\circ}$. After walking $50$ m closer to the tower to point $B$, the angle of elevation is $40^{\circ}$. Find the distance from $B$ to the top of the tower $T$.

In $\triangle ABT$: $\angle TAB = 25^{\circ}$ $\angle TBA = 180^{\circ} - 40^{\circ} = 140^{\circ}$ $\angle ATB = 180^{\circ} - (25^{\circ} + 140^{\circ}) = 15^{\circ}$ Using Sine Rule in $\triangle ABT$ to find $BT$ (let $BT = x$): $\frac{x}{\sin 25^{\circ}} = \frac{50}{\sin 15^{\circ}}$ $x = \frac{50 \times \sin 25^{\circ}}{\sin 15^{\circ}}$ $x \approx \frac{50 \times 0.4226}{0.2588} \approx 81.6 \text{ m}$

First, we identify the angles inside the triangle formed by the two observation points and the top of the tower. We use the properties of angles on a straight line and the sum of angles in a triangle. Then, we apply the Sine Rule to find the required distance.

1. Find the third angle $Z$: $Z = 180^{\circ} - (110^{\circ} + 35^{\circ}) = 35^{\circ}$
2. Use the Sine Rule to find side $z$: $\frac{z}{\sin Z} = \frac{x}{\sin X}$ $\frac{z}{\sin 35^{\circ}} = \frac{15}{\sin 110^{\circ}}$ $z = \frac{15 \times \sin 35^{\circ}}{\sin 110^{\circ}}$ $z \approx 9.16 \text{ cm}$ (Note: Since angle $Y =$ angle $Z$, this is an isosceles triangle, so $y = z = 9.16$ cm).

To find a side, we first ensure we have the angle opposite that side. Using the sum of angles in a triangle, we calculate the missing angle and then apply the Sine Rule ratio.

1. Determine the internal angles of triangle $L_1 L_2 S$: Angle at $L_1 = 90^{\circ} - 40^{\circ} = 50^{\circ}$ Angle at $L_2 = 310^{\circ} - 270^{\circ} = 40^{\circ}$
2. Calculate the angle at the ship $S$: $S = 180^{\circ} - (50^{\circ} + 40^{\circ}) = 90^{\circ}$
3. Use the Sine Rule to find distance $L_2 S$ (side $l_1$): $\frac{l_1}{\sin 50^{\circ}} = \frac{20}{\sin 90^{\circ}}$ $l_1 = 20 \times \sin 50^{\circ}$ $l_1 \approx 15.3 \text{ km}$

Convert bearings into internal triangle angles by referencing the horizontal (East-West) line. Once two angles and the side between them are known, the Sine Rule can solve for the remaining distances.