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Circle theorems 2: Questions

November 24, 2010 Leave a comment

Set 1

Set 2

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Categories: Geometry

Circle Theorems 2

November 23, 2010 Leave a comment

Tangents to circles

Investigation

Theorem 1

The length of the tangents from a point to a circle are of equal length

Theorem 2

The Angle between a tangent and a radius is a right-angle

Theorem 3

The angel between a tangent and a cord is equal to the angle subtended by the cord

Examples

Given that PA is a tangent find the value of x
Since PA is a tangent \angle PAO = 90^o
\implies x = 180^o-90^o-65^o=25^o \because 180^o \mbox{in a }\triangle

Given that PA and PB are tangents find the value of x
Since PA and PB are tangents they must both be of equal length.  Therefore \triangle ABP is isosceles.
\implies \angle ABP =\frac{180^o-49^o}{2}=65.5^o \because base angles in an isosceles triangle are equal.
\implies x=90^o-65.5^o=24.5^o \because the angle between a tangent and a radius is a right-angle.

Find the value of x
Since \triangle ABC \mbox{ is isosceles}\implies \angle ACB = 50^o
\implies x=50^o \because \mbox{Angle subtended by the cord}

Questions

Categories: Geometry

Trigonometry: Right-angled triangles Quesitons

November 22, 2010 Leave a comment

For each set of questions find the value of \mbox{w, x, y and z}

Make sure to write down all the working.  It is a good habit to be rigorous in your answers.

Set 1


Set 2

 


Set 3


Set 4


Set 5

Categories: Triangles

Trigonometry and Right-angled triangles

November 21, 2010 Leave a comment

Conceptual Understanding

If we find the ratio between a pair of sides on a triangle the value will be a constant even on larger triangles provided the triangles are similar to each other.

The right-angled trigonometry triangle

The hypotenuse is the longest side and is opposite the right-angle

The opposite is the side that is opposite the given angle. (Or the angle we wish to find)

The adjacent is the side next to the given angle.  (It is between the given angle and the right angle)

The ratios SohCahToa

 

\displaystyle sin \theta = \frac{opp}{hyp}

\displaystyle cos \theta = \frac{adj}{hyp}

\displaystyle tan \theta = \frac{opp}{adj}

When solving a trigonometry right-angled triangle problem we need two pieces of information and then use the relevant ratio to find the third.

There are three possible scenarios.

  1. We are finding an angle
  2. We are finding a side that is a numerator in the ratio
  3. We are finding a side that is a denominator in the ratio

Examples

Type 1 (The unknown is the numerator)

We have the hypotenuse and we want the opposite side.  The only ratio with hypotenuse and opposite in it is the sine ratio.

\displaystyle sin\theta = \frac{opp}{hyp}

Let’s fill out the equation

\displaystyle sin 35^o=\frac{x}{17}

\displaystyle \implies 0.374=\frac{x}{17}

\implies x=9.75cm


Type 2 (The unknown is the denominator) 

We have the opposite and we want the adjacent side.  The only ratio with adjacent and opposite in it is the tangent ratio.

\displaystyle tan \theta = \frac{opp}{adj}

Let’s fill out the equation

\displaystyle tan 27^o=\frac{28}{x}

\implies x\times tan 27^o=28

\displaystyle \implies x=\frac{28}{tan 27^o}

\therefore x =54.95mm


Type 3 (The angle is unknown) 

We have the hypotenuse and adjacent side.  The only ratio with adjacent and hypotenuse in it is the cosine ratio.

\displaystyle cos \theta = \frac{adj}{hyp}

Let’s fill out the equation

\displaystyle cos x=\frac{15}{32}

\implies \displaystyle x=cos^{-1}\left(\frac{15}{32}\right)

\implies x =62.0^o

Questions

Categories: Triangles

Pythagoras’ Theorem Questions

November 20, 2010 Leave a comment

Set 1: Find the value of a, b, c and d

Set 2: Find the value of a, b, c and d

Set 3: Find the value of a and c in the first diagram and the area of the second diagram

Categories: Triangles

Pythagoras’ Theorem Proof

November 20, 2010 Leave a comment

Consider a pair of right-angled similar triangles

Consider \triangle ABD

Since it is right-angled that implies that \beta + \delta = 90^o

Which implies that \triangle ACD is similar to the other two triangle.

Since the triangles are similar

\displaystyle \frac{a}{c}=\frac{c}{a+d}

\implies a^2+ad=c^2

Also \displaystyle \frac{b}{d}=\frac{a}{b}

\implies ad=b^2

Substituting gives \displaystyle a^2+b^2=c^2

Categories: Geometry

Angle at the centre of a circle proof

November 19, 2010 Leave a comment

Proof

AO=BO=CO=DO \because \mbox{they are all radii of a circle}

\implies \triangle AOC \mbox{ and }\triangle BOC \mbox{are isosceles}

Let \angle OAC=\alpha

\implies \angle ACO=\alpha\because \mbox{base angles in an isosceles trinagle are equal}

\mbox{So } \angle COA = 180^o-2\alpha\because 180^o\mbox{ in a triangle}

\implies \angle AOD = 2\alpha \because 180^o\mbox{ on a line}

Let \angle CBO=\beta

\implies \angle OCB=\beta\because \mbox{base angles in an isosceles trinagle are equal}

\mbox{So } \angle BOC= 180^o-2\beta\because 180^o\mbox{ in a triangle}

\implies \angle DOB = 2\beta\because 180^o\mbox{ on a line}

\mbox{So } \angle ACB = (\alpha + \beta)

\mbox{and } \angle AOB=(2\alpha + 2\beta)=2(\alpha + \beta)

\implies \angle AOB=2(\angle ACB)

Q.E.D.

Categories: Geometry