Calculate the size of angle ACB.

Calculate the area of the building’s footprint, ABC.

Calculate the volume of the office block.

To stabilize the structure, a steel beam must be made that runs from point C to point Q.

Calculate the length of CQ.

Calculate the angle CQ makes with BC.

$\cos {\text{ACB}} = \frac{{{{30}^2} + {{50}^2} - {{70}^2}}}{{2 \times 30 \times 50}}$     (M1)(A1)

Note: Award (M1) for substituted cosine rule formula, (A1) for correct substitution.

${\text{ACB}} = {120^ \circ }$     (A1)(G2)

${\text{Area of triangle ABC}} = \frac{{30(50)\sin {{120}^ \circ }}}{2}$     (M1)(A1)(ft)

Note: Award (M1) for substituted area formula, (A1)(ft) for correct substitution.

$= 650{\text{ }}{{\text{m}}^2}$ $(649.519 \ldots {\text{ }}{{\text{m}}^2})$     (A1)(ft)(G2)

Notes: The answer is $650{\text{ }}{{\text{m}}^2}$ ; the units are required. Follow through from their answer in part (a).

${\text{Volume}} = 649.519 \ldots \times 120$     (M1)
$= 77900{\text{ }}{{\text{m}}^3}$ ($77942.2 \ldots {\text{ }}{{\text{m}}^3}$)     (A1)(G2)

Note: The answer is $77900{\text{ }}{{\text{m}}^3}$ ; the units are required. Do not penalise lack of units if already penalized in part (b). Accept $78000{\text{ }}{{\text{m}}^3}$ from use of 3sf answer $650{\text{ }}{{\text{m}}^2}$ from part (b).

${\text{C}}{{\text{Q}}^2} = {50^2} + {120^2}$     (M1)
${\text{CQ}} = 130{\text{ (m)}}$     (A1)(G2)

Note: The units are not required.

$\tan {\text{QCB}} = \frac{{120}}{{50}}$     (M1)

Note: Award (M1) for correct substituted trig formula.

${\text{QCB}} = {67.4^ \circ }$ ($67.3801 \ldots$)     (A1)(G2)

Note: Accept equivalent methods.