Trigonometry is closely related to triangles. There are several types of Trigonometric Comparisons in Right Triangles, namely sin, cos, tan.
Hello Friends! This time we will learn about trigonometric ratios in right triangles. Let’s see more in here.
Trigonometry is closely related to triangles. There are several types of trigonometric comparisons, namely sin, cos, tan, secan (sec), cossec (csc), and cotangent (cot). Trigonometry (from Greek trigonon “triangle” + metron “measure”).
Trigonometry helps us find angles and distances, and is widely used in science, engineering, video games, and more. We encounter many things related to trigonometric comparisons. Trigonometric comparisons can be used to solve contextual problems related to angles and distances. By studying trigonometric ratios we can
train reasoning in solving these problems.
Let’s start by remembering the right triangle first…
Right triangle
Look at the following right triangle image!
Right angles are indicated by small squares and suppose other angles are marked α. So some of the side terms are:
adjacent is the side adjacent (next to) angle α,
opposite which is the side opposite the corner α,
hypotenuse is the longest side of a right triangle.
Remember the Pythagorean theorem??? Yeahh… we will also use that in studying trigonometry…
Theorem of Pythagoras
In a right triangle, the hypotenuse is equal to the sum of the squares of the angled sides.
In the right triangle image above, this applies
c2=a2+b2 or c=a2+b2
to find the length of the hypotenuse or longest side or hypotenuse if the length of the right angle is known
a2=c2−b2 or a=c2−b2
to find the length of the front side if the lengths of the hypotenuse and side sides are known
b2=c2−a2 or b=c2−a2
to find the length of the side if the lengths of the hypotenuse and front side are known
Example of a Pythagorean question
Determine the value of x in the following triangle.
Solution
x is not the longest side so you can use formula 2 or 3
x=72−22x=49−4x=45
So, the value of x is 45.
Trigonometric Comparisons in Right Triangles
Look at the following right triangle,
Here are the Trigonometric Comparisons:
Sinus sinA=hypotenuseopposite=BABC=ca
Cosine
cosA=hypotenuseadjacent=BACA=cb
Tangen
tanA=adjacentopposite=CABC=ba
Secan
secA=cosA1=cb1=bc
Cosecan cscA=sinA1=ca1=ac
Cotangen cotA=tanA1=ba1=ab
Note: Terms that often appear are sin, cos, tan… the easiest to memorize are SOH, CAH, TOA..

Example of Trigonometry Comparison Questions
Determine the trigonometric ratio values sinθ,cosθ,tanθ in the following figure
Solution ✍️
find the length of side AC or side side with pythagoras
ACAC=AB2−BC2=102−82=100−64=36=6
mencari nilai sinθ,cosθ,tanθsinθ=mide=ABBC=108=54cosθ=misa=ABAC=106=53tanθ=sade=ACBC=68=34
It is known that the triangle PQR is right-angled at Q with side length PR=5 units and side length QR=2 units. If ∠RPQ=α, determine the values of sinα,cosα,tanα!
Solution ✍️
draw a right triangle and
Look at the following picture of a right triangle
find the length of side PQ or side side with Pythagoras.
PQPQ=PR2−RQ2=(5)2−22=5−4=1=1
looks for the values sinα,cosα,tanαsinα=mide=PRRQ=52=525cosα=misa=PRPQ=51=515tanα=sade=PQRQ=12=2
Given cosβ∘=21 and β acute angle (0∘<β<90∘). Find the value of the other trigonometric ratio of angle β.
Solution ✍️
identify the sides of the triangle
cosβ∘=21=misa, we get the length of side 1 and the length of hypotenuse 2
draw a right triangle, for example △ABC is right angled at B and ∠BAC=β then AB=1 and AC=2
Look at the following image
find the length of side AB or front side using Pythagoras.
BCBC=AC2−AB2=22−12=4−1=3
looks for the values sinβ,cosβ,tanβsinβ=mide=ACBC=23=213cosβ=misa=ACAB=21tanβ=sade=ABBC=13=3
Given tanα∘=43 and α acute angle (0∘<α<90∘). determine the value of sin2α+cos2αsinα−tanα.
Solution ✍️
identify the sides of the triangle
tanα∘=43=sade, we get the length of the front side is 3 and the length of the side is 4
draw a right triangle, for example △ABC is right angled at B and ∠BAC=α then AB=4 and BC=3
Look at the following image
find the length of side AC or hypotenuse using Pythagoras.
ACAC=AB2+BC2=32+42=9+16=25=5
finding value sin2α+cos2αsinα−tanαsin2α+cos2αsinα−tanα=(53)2+(54)253−43=259+25162012−2015=2525−203=−203
So, value sin2α+cos2αsinα−tanα=−203