In triangle ABC, BC = a , AC = b , AB = c and [BD] is perpendicular to [AC].

 

(a)     Show that ${\text{CD}} = b - c\cos A$.

(b)     Hence, by using Pythagoras’ Theorem in the triangle BCD, prove the cosine rule for the triangle ABC.

(c)     If ${\rm{A\hat BC}} = 60^\circ$ , use the cosine rule to show that $c = \frac{1}{2}a \pm \sqrt {{b^2} - \frac{3}{4}{a^2}}$ .

The above three dimensional diagram shows the points P and Q which are respectively west and south-west of the base R of a vertical flagpole RS on horizontal ground. The angles of elevation of the top S of the flagpole from P and Q are respectively 25° and 40° , and PQ = 20 m .

Determine the height of the flagpole.

(a)     ${\text{CD}} = {\text{AC}} - {\text{AD}} = b - c\cos A$     R1AG

[1 mark]

 

(b)     METHOD 1

${\text{B}}{{\text{C}}^2} = {\text{B}}{{\text{D}}^2} + {\text{C}}{{\text{D}}^2}$     (M1)

${a^2} = {(c\sin A)^2} + {(b - c\cos A)^2}$     (A1)

$= {c^2}{\sin ^2}A + {b^2} - 2bc\cos A + {c^2}{\cos ^2}A$     A1

$= {b^2} + {c^2} - 2bc\cos A$     A1

[4 marks]

METHOD 2

${\text{B}}{{\text{D}}^2} = {\text{A}}{{\text{B}}^2} - {\text{A}}{{\text{D}}^2} = {\text{B}}{{\text{C}}^2} - {\text{C}}{{\text{D}}^2}$     (M1)(A1)

$\Rightarrow {c^2} - {c^2}{\cos ^2}A = {a^2} - {b^2} + 2bc\cos A - {c^2}{\cos ^2}A$     A1

$\Rightarrow {a^2} = {b^2} + {c^2} - 2bc\cos A$     A1

[4 marks]

 

(c)     METHOD 1

${b^2} = {a^2} + {c^2} - 2ac\cos 60^\circ  \Rightarrow {b^2} = {a^2} + {c^2} - ac$     (M1)A1

$\Rightarrow {c^2} - ac + {a^2} - {b^2} = 0$     M1

$\Rightarrow c = \frac{{a \pm \sqrt {{{( - a)}^2} - 4({a^2} - {b^2})} }}{2}$     (M1)A1

$= \frac{{a \pm \sqrt {4{b^2} - 3{a^2}} }}{2} = \frac{a}{2} \pm \sqrt {\frac{{4{b^2} - 3{a^2}}}{4}}$     (M1)A1

$= \frac{1}{2}a \pm \sqrt {{b^2} - \frac{3}{4}{a^2}}$     AG

Note: Candidates can only obtain a maximum of the first three marks if they verify that the answer given in the question satisfies the equation.

 

[7 marks]

METHOD 2

${b^2} = {a^2} + {c^2} - 2ac\cos 60^\circ  \Rightarrow {b^2} = {a^2} + {c^2} - ac$     (M1)A1

${c^2} - ac = {b^2} - {a^2}$     (M1)

${c^2} - ac + {\left( {\frac{a}{2}} \right)^2} = {b^2} - {a^2} + {\left( {\frac{a}{2}} \right)^2}$     M1A1

${\left( {c - \frac{a}{2}} \right)^2} = {b^2} - \frac{3}{4}{a^2}$     (A1)

$c - \frac{a}{2} =  \pm \sqrt {{b^2} - \frac{3}{4}{a^2}}$     A1

$\Rightarrow c = \frac{1}{2}a \pm \sqrt {{b^2} - \frac{3}{4}{a^2}}$     AG

[7 marks]

${\text{PR}} = h\tan 55^\circ {\text{ , QR}} = h\tan 50^\circ {\text{ where RS}} = h$     M1A1A1

Use the cosine rule in triangle PQR.     (M1)

${20^2} = {h^2}{\tan ^2}55^\circ  + {h^2}{\tan ^2}50^\circ  - 2h\tan 55^\circ h\tan 50^\circ \cos 45^\circ$     A1

${h^2} = \frac{{400}}{{{{\tan }^2}55^\circ  + {{\tan }^2}50^\circ  - 2\tan 55^\circ \tan 50^\circ \cos 45^\circ }}$     (A1)

$= 379.9…$     (A1)

$h = 19.5{\text{ (m)}}$     A1

[8 marks]

The majority of the candidates attempted part A of this question. Parts (a) and (b) were answered reasonably well. In part (c), many candidates scored the first two marks, but failed to recognize that the result was a quadratic equation, and hence did not progress further.

Correct answers to part B were rarely seen. Although many candidates expressed RS correctly in two different ways, they failed to go on to use the cosine rule.