Right-angled triangles

1 Right-angled triangles

Look at Figure 1 which could, for example, be a profile of a hill with a constant gradient.

Figure 1

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The two right-angled triangles $A B_{1} C_{1}$ and $A B_{2} C_{2}$ are similar (because the three angles of triangle $A B_{1} C_{1}$ are equal to the equivalent 3 angles of triangle $A B_{2} C_{2}$ ). From the basic properties of similar triangles corresponding sides have the same ratio. Thus, for example,

$\frac{B_{1} C_{1}}{A B_{1}} = \frac{B_{2} C_{2}}{A B_{2}} and \frac{A C_{1}}{A B_{1}} = \frac{A C_{2}}{A B_{2}} (1)$

The values of the two ratios (1) will clearly depend on the angle $A$ of inclination. These ratios are called the sine and cosine of the angle $A$ , these being abbreviated to $sin A$ and $cos A$ .

Key Point 1

Figure 2

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$A C$ is the side adjacent to angle $A$ .

$B C$ is the side opposite to angle $A$ .

$A B$ is the hypotenuse of the triangle (the longest side).

Task!

Referring again to Figure 2 in Key Point 1, write down the ratios which give $sin B$ and $cos B$ .

$sin B = \frac{A C}{A B} cos B = \frac{B C}{A B}$ .

Note that $sin B = cos A = cos (9 0^{\circ} - B)$ and $cos B = sin A = sin (9 0^{\circ} - B)$

A third result of importance from Figure 1 is

$\frac{B_{1} C_{1}}{A C_{1}} = \frac{B_{2} C_{2}}{A C_{2}}$ (2)

These ratios is referred to as the tangent of the angle at $A$ , written $tan A$ .

Key Point 2

Figure 3

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$tan A = \frac{B C}{A C} = \frac{length of opposite side}{length of adjacent side}$

For any right-angled triangle the values of sine, cosine and tangent are given in Key Point 3.

Key Point 3

Figure 4

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We can write, therefore, for any right-angled triangle containing an angle $θ$ (not the right-angle)

\begin{array}{rcl} sin θ & = & \frac{length of side opposite angle θ}{length of hypotenuse} = \frac{Opp}{Hyp} \\ cos θ & = & \frac{length of side adjacent to angle θ}{length of hypotenuse} = \frac{Adj}{Hyp} \\ tan θ & = & \frac{length of side opposite angle θ}{length of side adjacent to angle θ} = \frac{Opp}{Adj} \end{array}

These are sometimes memorised as $S O H$ , $C A H$ and $T O A$ respectively.

These three ratios are called trigonometric ratios .

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Solution

By Pythagoras’ theorem ${(A B)}^{2} = x^{2} + x^{2} = 2 x^{2}$ so $A B = x \sqrt{2}$

Hence $sin 4 5^{\circ} = \frac{B C}{A B} = \frac{x}{x \sqrt{2}} = \frac{1}{\sqrt{2}} cos 4 5^{\circ} = \frac{A C}{A B} = \frac{1}{\sqrt{2}} tan 4 5^{\circ} = \frac{B C}{A C} = \frac{x}{x} = 1$

1 Right-angled triangles

Key Point 1

Task!

Key Point 2

Key Point 3

Task!

Key Point 4

Example 1

Solution

1 Right-angled triangles

Answer

Answer