Precalculus - Right Triangle Trigonometry

Right Triangles [Recap]

• A right triangle is a special type of triangle in which the measure of one angle is exactly $90°$. The measure of the other two angles sums up to $90°$.
• Each side of a right triangle has a name, namely - hypotenuse, altitude (perpendicular), and base.
• The largest side of any right triangle is termed a hypotenuse and is always opposite to the right angle.
• The other two sides forming the right angle are called base and altitude (perpendicular).
• A right triangle is shown in the figure below. 

• In the given right triangle $ABC$, side $AC$ opposite to the right angle is the hypotenuse.
• Notice that the altitude of a right triangle is defined relative to an angle. For example, relative to $\angle C$ side AB acts as altitude. Relative to $\angle A$, side BC acts as altitude.
• The side left after assigning hypotenuse, and altitude is designated as a base.
• A right triangle obeys Pythagoras' theorem i.e ${\left(hypotenuse\right)}^2={\left(base\right)}^2+{\left(altitude\right)}^2$.

Trigonometric Ratios

• In a right-angled triangle, the ratios of sides concerning any of its acute angles are known as trigonometric ratios of that particular angle.
• In total, there are six trigonometric ratios.
• Consider right triangle ABC (as shown in the figure below) right angled at B, the trigonometric ratios of $\angle A$ are defined as follows :

sine of $\angle A =\frac{altitude}{hypotenuse}$$=\frac{BC}{AC}$

cosine of $\angle A =\frac{base}{hypotenuse}$$=\frac{AB}{AC}$

tangent of $\angle A =\frac{altitude}{base}$$=\frac{BC}{AB}$

cotangent of $\angle A =\frac{base}{altitude}$$=\frac{AB}{BC}$

secant of $\angle A =\frac{hypotenuse}{base}$$=\frac{AC}{AB}$

cosecant of $\angle A =\frac{hypotenuse}{altitude}$$=\frac{AC}{BC}$

Trigonometric ratios of complementary angles

• A pair of angles whose sum is $90°$ called complementary angles.
• In a right-angled triangle, the two acute angles are always complementary.
• Thus, for a right-angled triangle, as shown in the figure:

$\sin \theta =\frac{BC}{AC}= \cos \left(90°-\theta \right)$ 

$\cos \theta =\frac{AB}{AC}= \sin \left(90°-\theta \right)$

$tan \theta =\frac{BC}{AB}= \mathrm{co}t \left(90°-\theta \right)$

$cot \theta =\frac{AB}{BC}= \tan \left(90°-\theta \right)$

$sec \theta =\frac{AC}{AB}= \cos ec \left(90°-\theta \right)$

$cosec \theta =\frac{AC}{BC}= sec \left(90°-\theta \right)$

Solved Examples

Example 1: Find the values of $\sin \theta$, $\cos \theta$ and $\tan \theta$ for the right triangle as shown in the figure.

Solution: For the angle '$\theta$', $hypotenuse = PQ=10 cm$, $altitude=QR=8 cm$ and $base = PR = 6 cm$. Thus, 

$\sin \theta =\frac{altitude}{hypotenuse}$$=\frac{8}{10}=\frac{4}{5}$

$\cos \theta =\frac{base}{hypotenuse}$$=\frac{6}{10}=\frac{3}{5}$

$\tan \theta =\frac{altitude}{base}$$=\frac{8}{6}=\frac{4}{3}$

Example 2: In a $△ABC$, right angled at B, if $\sin A=\frac{1}{\sqrt{3}}$, then find the value of $\sin A·\cos A·\tan A$.

Solution: It is given that $\sin A=\frac{1}{\sqrt{3}}$ but $\sin A=\frac{altitude}{hypotenuse}$. From here, we can conclude that $altitude = k$ and $hypotenuse = \sqrt{3}k$, where $k$ is any positive number. Let us now draw the triangle with the known data.

Using the Pythagorean theorem, we get, 

     ${\left(base\right)}^2={\left(AB\right)}^2={\left(k\sqrt{3}\right)}^2-{\left(k\right)}^2$

or,  ${\left(base\right)}^2={\left(AB\right)}^2=3k^2-k^2=2k^2$

so,  $base=AB=k\sqrt{2}$

Now, $cos A=\frac{base}{hypotenuse}=\frac{AB}{AC}$$=\frac{k\sqrt{2}}{k\sqrt{3}}=\frac{\sqrt{2}}{\sqrt{3}}$

similarly, $\tan A=\frac{altitude}{base}=\frac{BC}{AB}$$=\frac{k}{k\sqrt{2}}=\frac{1}{\sqrt{2}}$

Therefore, $\sin A·\cos A·\tan A=$$\frac{1}{\sqrt{3}}\times \frac{\sqrt{2}}{\sqrt{3}}\times \frac{1}{\sqrt{2}}$$=\frac{1}{3}$

Example 3: In a right triangle if $\sin \left(\theta +30°\right)=\cos 2\theta$, find the value of the acute angle $\theta$.

Solution: It is given that $\sin \left(\theta +30°\right)=\cos 2\theta$. We can rewrite this expression as $\sin \left(\theta +30°\right)=\sin \left(90°-2\theta \right)$.

Therefore, $\theta +30°=90°-2\theta$ $\Rightarrow 3\theta =90°-30°=60°$ $\Rightarrow \theta =\frac{60°}{3}=20°$

Example 4: Find the value of $\frac{\sin 40°}{\cos 50°}+\frac{\tan 20°}{cot 70°}-1$.

Solution: $\frac{\sin 40°}{\cos 50°}+\frac{\tan 20°}{cot 70°}-1$$=\frac{\sin 40°}{\sin \left(90°-50°\right)}+\frac{\tan 20°}{\tan \left(90°-70°\right)}-1$$=\frac{\sin 40°}{\sin 40°}+\frac{\tan 20°}{\tan 20°}-1$$=1+1-1=1$

Cheat Sheet

In any right-angled triangle, three basic trigonometric ratios defined for an acute angle '$\theta$' are:

• $\sin \theta =\frac{altitude}{hypotenuse}$
• $\cos \theta =\frac{base}{hypotenuse}$
• $\tan \theta =\frac{altitude}{base}$

Relationship between trigonometric ratios:

• $\sin \theta ·\cos ec \theta =1$  or  $\sin \theta =\frac{1}{\cos ec \theta }$  or  $\cos ec \theta =\frac{1}{\sin \theta }$
• $\cos \theta ·sec \theta =1$  or  $\cos \theta =\frac{1}{sec \theta }$  or  $sec \theta =\frac{1}{\cos \theta }$
• $\tan \theta ·cot \theta =1$  or  $\tan \theta =\frac{1}{cot \theta }$  or  $\mathrm{co}t \theta =\frac{1}{\tan \theta }$
• $\tan \theta =\frac{\sin \theta }{\cos \theta }$
• $cot \theta =\frac{\cos \theta }{\sin \theta }$

Blunder Areas

• A triangle can never have two right angles.
• Trigonometric ratios apply to right-angled triangles only.