16N.1.SL.TZ0.T_11

pestleMathematicsAASLPaper 116N· sl-3-2-2d-and-3d-trig-sine-rule-cosine-rule-areasource ↗

AC is a vertical communications tower with its base at C.

The tower has an observation deck, D, three quarters of the way to the top of the tower, A.

N16/5/MATSD/SP1/ENG/TZ0/11

From a point B, on horizontal ground 250 m from C, the angle of elevation of D is 48°.

Calculate CD, the height of the observation deck above the ground.

[2]

Calculate the angle of depression from A to B.

[4]

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\tan 48^\circ = \frac{{{\text{CD}}}}{{250}}$ (M1)

Note: Award (M1) for correct substitution into the tangent ratio.

$({\text{CD}} = ){\text{ }}278{\text{ }}({\text{m}}){\text{ }}(277.653 \ldots )$ (A1) (C2)

[2 marks]

$\tan {\text{ABC (or equivalent)}} = \frac{{\frac{4}{3} \times 277.653 \ldots }}{{250}}$ (M1)(M1)(M1)

Note: Award (M1) for $\frac{4}{3}$ multiplying their part (a), (M1) for substitution into the tangent ratio, (M1) for correct substitution.

$90 - {\tan ^{ - 1}}\left( {\frac{{250}}{{\frac{4}{3} \times 277.653 \ldots }}} \right)$ (M1)(M1)(M1)

Note: Award (M1) for $\frac{4}{3}$ multiplying their part (a), (M1) for substitution into the tangent ratio, (M1) for subtracting from 90 and for correct substitution.

${\text{(angle of depression}} = {\text{) }}56.0^\circ {\text{ }}(55.9687 \ldots )$ (A1)(ft) (C4)

Note: Follow through from part (a).

[4 marks]

Examiners’ report

[N/A]