It is given that \(\tan\theta=\frac{\sqrt5}{5}\) and \(180^\circ\lt \theta\lt 360^\circ\).

(a) Find the value of \(\cos\theta\).

(b) Find the value of \(\sin\theta\).

(c) Find the value of \(\sec\theta+\cot\theta\). Give your answer in the form \(\frac{a+b\sqrt c}{\sqrt5}\), where \(a\), \(b\) and \(c\) are integers.

Do not use a calculator in this question.

The diagram shows triangle \(ABC\), where \(AB=\sqrt2\), \(\angle A=60^\circ\), \(\angle B=75^\circ\) and \(\angle C=45^\circ\).

You may use \(\sin60^\circ=\frac{\sqrt3}{2}\), \(\sin45^\circ=\frac{\sqrt2}{2}\), \(\cos60^\circ=\frac12\), \(\cos45^\circ=\frac{\sqrt2}{2}\), \(\tan60^\circ=\sqrt3\) and \(\tan45^\circ=1\).

(a) Given that the area of triangle \(ABC\) is \(\frac{3+\sqrt3}{4}\), show that

\(\sin75^\circ=\frac{\sqrt6+\sqrt2}{4}.\)

(b) Hence find the exact length of \(AC\).

This question is to be answered without using a calculator.

(a) In triangle \(ABC\), \(AB=5\sqrt3-6\), \(BC=5\sqrt3+6\) and angle \(ABC=120^\circ\). Find \(AC\) in the form \(a\sqrt b\), where \(a\) and \(b\) are integers.

(b) In triangle \(PQR\), \(PQ=3+2\sqrt5\), angle \(PQR=30^\circ\) and the area of triangle \(PQR\) is \(\frac14(2+5\sqrt5)\). Find \(QR\) in the form \(c+d\sqrt5\), where \(c\) and \(d\) are integers.

(a) Find the exact coordinates of the points of intersection of \(y=x^2+2\sqrt5x-20\) and \(y=3\sqrt5x+10\).

(b) Given \(\tan\theta=\frac{\sqrt3-1}{2+\sqrt3}\), for \(0\lt \theta\lt \frac\pi2\), find \(\operatorname{cosec}^2\theta\) in the form \(a+b\sqrt3\).

Given that \(x=\operatorname{sec}^{2}\theta\) and \(y+2=\operatorname{cot}^{2}\theta\), find \(y\) in terms of \(x\).

In this question all lengths are in centimetres.

The diagram shows triangle \(ABC\) with \(AB=\sqrt5-1\), \(AC=\sqrt5+1\) and \(BC=x\). The area of triangle \(ABC\) is \(\frac{2\sqrt5}{3}\text{ cm}^2\). Angle \(A\) is acute.

(a) Find the exact value of \(\sin A\).

(b) Find the exact value of \(\cos A\) and hence find the exact value of \(x\).

(c) Find the exact value of \(\sin B\).

The diagram shows triangle \(ABC\) with

\(AC=\sqrt6-\sqrt2,\quad AB=\sqrt6+\sqrt2\)

and angle \(CAB=60^\circ\).

(a) Find the exact length of \(BC\).

(b) Show that

\(\sin ACB=\frac{\sqrt6+\sqrt2}{4}.\)

(c) Show that the perpendicular distance from \(A\) to the line \(BC\) is \(1\).

The diagram shows a trapezium \(ABCD\), where \(AB\) is parallel to \(DC\), \(AD\) is perpendicular to \(DC\),

\(AB=2+3\sqrt5,\qquad DC=6+3\sqrt5,\qquad AD=10-2\sqrt5.\)

(i) Find the area of the trapezium in the form \(a+b\sqrt5\), where \(a\) and \(b\) are integers.

(ii) Find \(\operatorname{cot} BCD\) in the form \(c+d\sqrt5\), where \(c\) and \(d\) are constants.

Do not use a calculator in this question. In triangle \(ABC\), \(AB=2\sqrt5-1\), \(BC=2+\sqrt5\), and angle \(ABC=90^\circ\).

(i) Find the exact length of \(AC\).

(ii) Find \(\tan ACB\), giving your answer in the form \(p+q\sqrt r\), where \(p\), \(q\) and \(r\) are integers.

(iii) Hence find \(\operatorname{sec}^2ACB\), giving your answer in the form \(s+t\sqrt u\), where \(s\), \(t\) and \(u\) are integers.