Trigonometry: Finding Trigonometric Ratios (Exercise 8.1)

This lesson covers worked questions from Exercise 8.1 on right-angled triangle trigonometry. In each case we use the given information and the Pythagoras theorem to fix the triangle, then read off the required ratios.

All ratios from $\sin A = \tfrac{4}{5}$

We are given $\sin A = \tfrac{4}{5}$, which is the opposite side over the hypotenuse. In right triangle $ABC$ with the right angle at $B$, take the side opposite $A$ as $4k$ and the hypotenuse as $5k$.

By the Pythagoras theorem the remaining side $AB$ is

$AB = \sqrt{(5k)^2 - (4k)^2} = \sqrt{25k^2 - 16k^2} = \sqrt{9k^2} = 3k.$

Now every ratio follows:

$\sin A = \tfrac{4}{5}, \quad \csc A = \tfrac{5}{4},$ $\cos A = \tfrac{3}{5}, \quad \sec A = \tfrac{5}{3},$ $\tan A = \tfrac{4}{3}, \quad \cot A = \tfrac{3}{4}.$

Showing $2\sin A \cos A = 1$ when $\tan A = 1$

In right triangle $ABC$ with $\angle B = 90^\circ$, suppose $\tan A = 1$. Since $\tan A = \tfrac{BC}{AB} = 1$, the two legs are equal: $BC = AB = k$.

The hypotenuse is

$AC = \sqrt{k^2 + k^2} = \sqrt{2k^2} = \sqrt{2}\,k.$

So

$\sin A = \frac{k}{\sqrt{2}\,k} = \frac{1}{\sqrt{2}}, \qquad \cos A = \frac{k}{\sqrt{2}\,k} = \frac{1}{\sqrt{2}}.$

Therefore

$2\sin A \cos A = 2 \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = 2 \cdot \frac{1}{2} = 1.$

Triangle $OPQ$: finding $\sin Q$ and $\cos Q$

In right triangle $OPQ$ with $\angle P = 90^\circ$, we are given $OP = 7$ cm and $OQ - PQ = 1$ cm, so $OQ = 1 + PQ$.

By the Pythagoras theorem $OQ^2 = OP^2 + PQ^2$. Substituting $OQ = 1 + PQ$,

$(1 + PQ)^2 = OP^2 + PQ^2.$

Expanding the left side with $(a+b)^2 = a^2 + 2ab + b^2$ and cancelling $PQ^2$,

$1 + 2\,PQ = OP^2 = 7^2 = 49.$

So $2\,PQ = 48$, giving $PQ = 24$ and $OQ = 1 + 24 = 25$.

Now

$\sin Q = \frac{OP}{OQ} = \frac{7}{25}, \qquad \cos Q = \frac{PQ}{OQ} = \frac{24}{25}.$

Verifying an identity when $\cot A = \tfrac{4}{3}$

We check whether $\dfrac{1 - \tan^2 A}{1 + \tan^2 A} = \cos^2 A - \sin^2 A$.

Given $\cot A = \tfrac{4}{3}$, so $\tan A = \tfrac{3}{4}$. Take the opposite side $3k$ and the adjacent side $4k$; then the hypotenuse is

$AC = \sqrt{(4k)^2 + (3k)^2} = \sqrt{25k^2} = 5k,$

so $\cos A = \tfrac{4}{5}$ and $\sin A = \tfrac{3}{5}$.

Left side

$\frac{1 - \left(\tfrac{3}{4}\right)^2}{1 + \left(\tfrac{3}{4}\right)^2} = \frac{1 - \tfrac{9}{16}}{1 + \tfrac{9}{16}} = \frac{\tfrac{7}{16}}{\tfrac{25}{16}} = \frac{7}{25}.$

Right side

$\cos^2 A - \sin^2 A = \left(\tfrac{4}{5}\right)^2 - \left(\tfrac{3}{5}\right)^2 = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}.$

Both sides equal $\tfrac{7}{25}$, so the identity holds.

Triangle $PQR$: finding $\sin P$, $\cos P$ and $\tan P$

In right triangle $PQR$ with $\angle Q = 90^\circ$, we are given $PR + QR = 25$ cm and $PQ = 5$ cm, so $PR = 25 - QR$.

With $PR$ as the hypotenuse, $PR^2 = PQ^2 + QR^2$. Substituting,

$(25 - QR)^2 = PQ^2 + QR^2.$

Expanding the left side and cancelling $QR^2$,

$625 - 50\,QR = 5^2 = 25,$

so $50\,QR = 600$, giving $QR = 12$ and $PR = 25 - 12 = 13$.

Now

$\sin P = \frac{QR}{PR} = \frac{12}{13}, \qquad \cos P = \frac{PQ}{PR} = \frac{5}{13}, \qquad \tan P = \frac{QR}{PQ} = \frac{12}{5}.$

Key takeaways

• From any one trigonometric ratio you can build the right triangle and read off all six ratios using the Pythagoras theorem.
• When two sides are linked by a sum or difference, substitute into the Pythagoras theorem and expand the square to solve for the unknown side.
• An identity is confirmed by computing the left and right sides separately and checking they match.