7.2 Triangles

The triangle is isosceles therefore \(x = y\) (\(\angle\)s opp equal sides).

\begin{align*} 180° & = 36° - 2x \qquad \text{(sum of }\angle \text{s} \text{ in }\triangle \text{)} \\ 2x & = 144° \\ \therefore x & = 72° = y \end{align*}

\(x\) is an exterior angle, therefore \(P\hat{N}O + O\hat{P}N = x\) (ext \(\angle\) of \(\triangle\)).

\begin{align*} x & = 30° + 68° \\ &= 98° \end{align*}

First find \(y\). \(y + 68° = 180°\) (\(\angle\)s on a str line). Therefore \(y = 112°\).

\(y\) is an exterior angle, therefore \(P\hat{N}O + O\hat{P}N = y\) (ext \(\angle\) of \(\triangle\)).

\begin{align*} 112° & = x + 68°\\ x & = 112° - 68° \\ & = 44° \end{align*}

Therefore \(y = 112°\) and \(x = 44°\).

\begin{align*} N\hat{P}O & = 180° - P\hat{N}O - N\hat{O}P \qquad \text{(sum of } \angle \text{s in } \triangle \text{)}\\ & = 180° - 90° - N\hat{O}P \\ & = 90° - N\hat{O}P \end{align*} \begin{align*} R\hat{S}O & = 180° - O\hat{R}S - R\hat{O}S \qquad \text{(sum of } \angle \text{s in } \triangle \text{)}\\ & = 180° - 90° - R\hat{O}S \\ & = 90° - R\hat{O}S \end{align*}

\(N\hat{O}P = R\hat{O}S\) (vert opp \(\angle\)s).

\(\therefore N\hat{P}O = R\hat{S}O\).

Therefore \(\triangle NPO\) and \(\triangle ROS\) are similar because they have the same angles.

Similar triangles have proportional sides:

\begin{align*} \therefore \frac{NP}{RS} &= \frac{NO}{OR}\\ \frac{19}{76}&=\frac{x}{116}\\ \therefore x &= 29 \end{align*}

From the theorem of Pythagoras we have:

\begin{align*} x^2 &= 15^2 + 20^2 \\ \therefore x &= \sqrt{625} \\ &= 25 \end{align*}

We note that:

\begin{align*} N\hat{P}O & = S\hat{R}T \qquad \text{(given)} \\ P\hat{N}O & = R\hat{T}S \qquad \text{(given)} \\ \therefore P\hat{N}O & = R\hat{T}S \qquad \text{(sum of} \angle \text{s in } \triangle \text{)} \\ \therefore \triangle NPO &\enspace|||\enspace \triangle TSR \text{ (AAA)} \end{align*}

Now we can use the fact that the sides are in proportion to find \(x\) and \(y\):

\begin{align*} \frac{NO}{OP} &= \frac{TS}{TR} \\ \frac{14}{12} &= \frac{21}{x} \\ x &= \frac{21 \times 12}{14} \\ &= 18 \\\\ \frac{OP}{NP} &= \frac{SR}{TR}\\ \frac{y}{12} & = \frac{6}{18} \\ 18y & = 72 \\ y & = 4 \end{align*}

Therefore \(x = 18\) and \(y = 4\).

From the theorem of Pythagoras:

\begin{align*} x^2 & = 15^2 - 9^2\\ x & = \sqrt{144}\\ & = 12 \\ \\ y^2 & = x^2 + 5^2 \\ y^2 & = 144 + 25 \\ y & = \sqrt{169} \\ y & = 13 \end{align*}

Therefore \(x = 12\) and \(y = 13\).

\begin{align*} AC & = CE \qquad \text{(given)} \\ BC & = CD \qquad \text{(given)} \\ A\hat{C}B & = D\hat{C}E \qquad \text{(vert opp } \angle \text{s} = \text{)} \\ \therefore \triangle ABC & \equiv \triangle EDC \qquad \text{SAS} \end{align*}

We have two equal sides (\(AB = BD\) and \(BC\) is common to both triangles) and one equal angle (\(\hat{A} = \hat{D}\)) but the sides do not include the known angle. The triangles therefore do not have a SAS and are therefore not congruent. (Note: \(A\hat{C}B\) is not necessarily equal to \(D\hat{C}B\) because it is not given that \(BC \perp AD\)).

There is not enough information given. We need at least three facts about the triangles and in this example we only know two sides in each triangle.

Note that \(BCD\) and \(ECA\) are not straight lines and so we cannot use vertically opposite angles.

There is not enough information given. Although we can work out which angles are equal we are not given any sides as equal. All we know is that we have two isosceles triangles. Note how this question differs from part a). In part a) we were given equal sides in both triangles, in this question we are only given that sides in the same triangle are equal.

\begin{align*} AC & = AC \qquad \text{(common side)} \\ B\hat{A}C & = D\hat{A}C \qquad \text{(given)} \\ A\hat{B}C & = A\hat{D}C \qquad \text{(given)} \\ \therefore \triangle ABC & \equiv \triangle ADC \qquad \text{AAS} \end{align*}