Right-Angled Triangles: Pythagoras and Trigonometry: past paper questions & worked solutions

Use Pythagoras because the 20 cm and 48 cm sides meet at the right angle.

Take the positive square root because x is a length.

The side of length 10 cm is the hypotenuse and x is adjacent to the given angle.

Multiply by 10 and round only the final value.

\(x\) is next to the \(65^\circ\) angle and \(10\) cm is the hypotenuse, so use cosine.

In the right-angled triangle, \(x\) is adjacent to the \(43^\circ\) angle.

The hypotenuse is \(6.5\text{ cm}\), and \(x\) is opposite the \(34^\circ\) angle.

In the right-angled triangle, \(x\) is adjacent to the \(42^\circ\) angle and \(9.5\text{ cm}\) is the hypotenuse.

Side \(AB\) is opposite the \(65^\circ\) angle, and \(AC=8.4\text{ cm}\) is the hypotenuse.

The vertical difference between the tops of the masts is \[ 30.7-6.2=24.5\text{ m} \]

The horizontal distance is \(244\text{ m}\).

The horizontal distance is \(12\text{ m}\) and the zip wire is \(13\text{ m}\).

Vertical difference: \[ \sqrt{13^2-12^2}=5 \]

Since the shorter post is \(2.6\text{ m}\), \[ AB=2.6+5=7.6 \]

The side \(6.5\) cm is opposite the \(42^\circ\) angle and \(x\) is the hypotenuse.

The vertical height is opposite the \(23^\circ\) angle, and \(AB=500\text{ m}\) is the hypotenuse.

Since \(D\) is on the positive \(x\)-axis and \(AD=17\),

First find \(AB\):

Angle \(ACB\):

Angle \(DCB\):

First find \(AB\):

Angle \(ACB\):

Angle \(DCB\):

Use the area to find \(AC\):

Then

Perimeter:

In right triangle \(BCD\),

In right triangle \(ABD\),

In triangle \(ABD\),

In triangle \(ABC\),

Since \(AC=BC\), the perpendicular support from \(C\) meets \(AB\) at its midpoint.

In right-angled triangle \(AMC\),

A whole number of metres must be bought, so buy \(7\) m for \(MC\).

Total length needed:

Cost:

In triangle \(ABC\), \[ BC^2=10^2-8^2 \]

So \[ EC=EB+BC=5+6=11 \]

In triangle \(ECD\), \[ w^2=11^2+14^2 \]

First find angle \(DAB\): \[ \cos(\angle DAB)=\frac{8}{14} \]

So \[ \angle CAD=55.150\ldots^\circ-38^\circ=17.150\ldots^\circ \]

In right-angled triangle \(ACD\), \[ \tan(17.150\ldots^\circ)=\frac{CD}{8} \]

Triangle \(AEB\) is right-angled:

For rectangle \(ABCD\),

Perimeter of the shaded shape:

Each triangle has hypotenuse \(16\text{ cm}\) and one side \(12.8\text{ cm}\).

Find the other side: \[ \sqrt{16^2-12.8^2}=9.6\text{ cm} \]

The exposed short vertical side is \[ 12.8-9.6=3.2\text{ cm} \]

Perimeter: \[ 3(16)+9.6+9.6+3.2=70.4 \]

Since \(ABC\) is isosceles and \(D\) is the midpoint of \(AC\), \(BD\) is perpendicular to \(AC\).

In right triangle \(ABD\), \[ BD=16,\qquad \angle DAB=65^\circ \]

Also, \[ \tan65^\circ=\frac{16}{AD} \]

In triangle \(DEF\), \[ EF=12\cos 40^\circ=9.192\ldots \]

In triangle \(EFG\), \[ \sin 28^\circ=\frac{EF}{EG} \]

All sides of the square have length equal to the square root of its area, and DC=EG.

Relative to angle E, FG is opposite and EG is adjacent.

The base EG and height FG are perpendicular.

In triangle \(ABC\), \[ AC^2=16^2+12^2=400 \]

Given \(AD=1.5AC\), \[ AD=30 \]

In triangle \(ABD\), \[ BD^2=30^2-16^2=644 \]

So \[ CD=BD-BC=25.377\ldots-12=13.377\ldots \]

In right-angled triangle \(AFD\), \[ AD^2=28^2+45^2 \]

Since \(AB=21\), \[ BD=53-21=32 \]

In right-angled triangle \(CBD\), \[ CB^2=35^2-32^2 \]

Area of triangle \(CDB\) is \[ \frac12\times BD\times CB \]

The height of the building is

The extra height from \(C\) to \(A\) is

So

For the angle of elevation \(\theta\) from \(D\),

Let the perpendicular from \(C\) meet \(AB\) at \(F\).

In right-angled triangle \(CFB\), \[ \tan x=\frac{8}{12.465\ldots} \]

A regular hexagon has interior angle \[ 120^\circ \]

In right-angled triangle \(ABP\), \[ \tan(\angle PAB)=\frac{BP}{AB}=\frac{2}{5} \]

Therefore \[ \angle PAF=120^\circ-21.801\ldots^\circ=98.198\ldots^\circ \]

So

(b) The angle of elevation to the top of the radio mast is \(25^\circ\).

In triangle \(ABC\),

In triangle \(DAC\), \(AD\) and \(AC\) are perpendicular.