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Length AB is 6 cm, length BG is 3 cm and length FG is 2 cm.

The length of the diagonal AF is 7 cm.

Calculate the angle between AF and the plane ABCD. Give the answer to 3 significant figures.

The plane ABCD is the base of the cuboid. The line FC and the plane ABCD form a right angle.

Draw the right-angled triangle AFC and label the sides. The angle between AF and the plane is \(x\).

Use: \(\sin{x} = \frac{o}{h}\)

\(\sin{x} = \frac{3}{7}\)

\(\sin{x} = 0.428571 \dotsc\) – do not round this answer yet.

To calculate the angle use the inverse sin button on the calculator \(\text{(}\sin^{-1}\text{)}\)

\(x = 25.4^\circ\)

Example

The plane ABCD is the base of the pyramid. The line VO and the plane ABCD form a right angle.

Draw the right-angled triangle OVC and label the sides. The angle between VC and the plane is (\(y\))

To find hypotenuse (Pythagoras) \( a^2 + b^2 = c^2\)

\( \text{CD}^2 + \text{AD}^2 = \text{AC}^2\)

\( 4^2 + 4^2 = c^2\)

\( 32 = c^2\)

\( c = \sqrt{32}\)

\( \sqrt{32}\) –  do not round this answer yet.

The length AC is \( \sqrt{32}\)

The point O is in the centre of the length AC so OC is half of the length AC.

The length OD is \( \frac{\sqrt{32}}{2}\) cm.

Use \(\tan{x} = \frac{o}{a}\)

\(\tan{x} = \frac{3}{\frac{\sqrt{32}}{2}}\)

\(\tan{x} = 1.06066 \dotsc\) – do not round this answer yet.

To calculate the angle use the inverse tan button on the calculator \(\text{(}\tan^{-1}\text{)}\)

\(x = 46.7^\circ\)

The sine rule is: \(\frac{a}{\sin{A}} = \frac{b}{\sin{B}} = \frac{c}{\sin{C}}\)

This version is used to calculate lengths.

It can be rearranged to: \( \frac{\sin{A}}{a} = \frac{\sin{B}}{b} = \frac{\sin{C}}{c}\)

This version is used to calculate angles.

Example

Calculate the angle PRQ. Give the answer to three significant figures.

Use the form \(\frac{\sin{A}}{a} = \frac{\sin{B}}{b} = \frac{\sin{C}}{c}\) to calculate the angle.

\(\frac{\sin{x}}{4} = \frac{\sin{75}}{9}\).

\(\sin{x} = 0.429300 \dotsc\) – do not round this answer yet.

To calculate the angle use the inverse sin button on the calculator

\(x = 25.4^\circ\).

The cosine rule is: \(a^2 = b^2 + c^2 - 2bc \cos{A}\).

This version is used to calculate lengths. \(\cos{A} = \frac{b^2 + c^2 - a^2}{2bc}\)

It can be rearranged to: \(\cos{A} = \frac{b^2 + c^2 - a^2}{2bc}\)

This version is used to calculate angles.

Example

Calculate the length BC. Give the answer to three significant figures.

Use the form: \( a^2 = b^2 + c^2 - 2bc \cos{A}\) to calculate the length.

\( \text{BC}^2 = 3^2 + 7^2 - 2 \times 3 \times 7 \cos{35}\)

\( \text{BC}^2 = 23.59561414 \dotsc\) – do not round this answer yet.

BC = 4.86 cm

Remember a bearing is an angle measured clockwise from north. The angle APB = \(200 - 165 = 35^\circ\).

Two sides and the angle in between are known. Calculate the length AB.

Use the cosine rule.

\(a^2 = b^2 + c^2 - 2bc \cos{A}\).

\( text{AB}^2 = 5.2^2 + 5.8^2-2 \times 5.2 \times 5.8 \cos{35}\).

\(\text{AB}^2 = 11.26874869\).

AB = 3.36 km

The area of any triangle can be calculated using the formula:

\(\text{Area of a triangle} = \frac{1}{2} ab \sin{C}\).

To calculate the area of any triangle the lengths of two sides and the angle in between are required.

Example

Calculate the area of the triangle. Give the answer to 3 significant figures.

Use the formula:

\(\text{area of a triangle} = \frac{1}{2} bc \sin{A}\)

\(\text{area} = \frac{1}{2} \times 7.1 \times 5.2 \sin{42}\)

area = \(12.4 \text{cm}^{2}\)

It may be necessary to rearrange the formula if the area of the triangle is given and a length or an angle is to be calculated.