Sine Rule, Cosine Rule and Area of a Triangle: past paper questions & worked solutions

Use the cosine rule for angle \(BAC\): \[ \cos A=\frac{8^2+9^2-10^2}{2(8)(9)} \]

Use the cosine rule for angle \(BAC\): \[ \cos A=\frac{8^2+9^2-10^2}{2(8)(9)} \]

A regular pentagon has interior angle \[ \frac{(5-2)\times180}{5}=108^\circ \]

Since \(BCE\) is a straight line, \[ \angle DCE=180^\circ-108^\circ=72^\circ \]

Area of triangle \(CDE\) is \[ \frac12ab\sin C \]

Use the area formula:

A kite can be split into two congruent triangles.

Area of triangle \(ABC\): \[ \frac12\times 6\times 11\times \sin 118^\circ \]

So the area of the kite is \[ 2\times \frac12\times 6\times 11\times \sin 118^\circ \]

The three triangle angles sum to 180 degrees.

Match each side to its opposite angle in the sine rule.

Use half the product of two sides times the sine of their included angle; retain unrounded values.

Round only the final area.

Split the isosceles triangle into two right-angled triangles.

The included angle is between the two given sides.

Take the positive square root for a length.

Use half the product of the two sides and the sine of their included angle.

First use triangle \(ABD\).

Use the cosine rule to find \(BD\):

In triangle \(BCD\),

Use the sine rule:

Correct to 3 significant figures,

First find \(BD\) in triangle \(ABD\) using the cosine rule.

In triangle \(BCD\), use the sine rule.

To 3 significant figures,

In triangle \(BCD\),

By the sine rule,

Now use cosine rule in triangle \(ABD\):

In triangle \(ABC\),

Use Heron's formula.

\(D\) is the foot of the perpendicular from \(B\) to \(AC\), so \(BD\) is the height of triangle \(ABC\).

In right-angled triangle \(BDF\),

Height of the parallelogram:

The horizontal offset from \(P\) to \(S\) is

So the horizontal distance from \(S\) to \(Q\) is

First use the cosine rule to find \(AC\): \[ AC^2=9.7^2+12.3^2-2(9.7)(12.3)\cos 115^\circ \]

Now use the sine rule: \[ \frac{\sin x}{9.7}=\frac{\sin 115^\circ}{18.607\ldots} \]

For triangle \(BCD\),

The angle \(BCD\) is acute, so

Find \(BD\):

Now use cosine rule in triangle \(ABD\):

Area of triangle \(ABC\): \[ \frac12(9)(12)\sin 60^\circ=46.765\ldots \]

Find \(AC\) using the cosine rule: \[ AC^2=9^2+12^2-2(9)(12)\cos 60^\circ \]

Area of triangle \(ACD\): \[ \frac12(10.816\ldots)(7)\sin 84^\circ=37.642\ldots \]

Total area: \[ 46.765\ldots+37.642\ldots=84.407\ldots \]

**Answer:** \(131\text{ cm}^2\).

(b) In triangle \(EFG\),

The two sides around the \(30^\circ\) angle are

Using the area formula,

Since \(\sin 30^\circ=\frac{1}{2}\),

So

Use the cosine rule to find \(BC\):

The largest side is \(AC=8\), so the largest angle is \(\angle B\).

Using the cosine rule again:

In triangle \(PQR\),

Use the sine rule:

Since angle \(PQR\) is obtuse,

Now find angle \(P\):

Area of triangle \(PQR\):

Correct to 3 significant figures:

In triangle \(BEC\), \(CE=13\) and \(\angle BEC=40^\circ\).

Since this is a prism, \(BE\) is perpendicular to the cross section, so triangle \(BEC\) is right-angled at \(B\).

The area of triangle \(ABC\) is \(22\text{ cm}^2\).

Using

The angle \(ACB\) is obtuse, so

Use the cosine rule:

Correct to one decimal place: