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Trigonometry and Geometry are the most important concepts in Mathematics. Both concepts while studied as separate chapters are somewhat interrelated. Geometry is related to all types of shapes in mathematics and trigonometry is about properties related to triangles. In addition, trigonometry mostly deals with the measurement of angles, and geometry is about concepts such as equality, congruence of triangles, and sum of angles.
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Key Takeaways: Trigonometry, Geometry, Pythagoras Theorem, Right-Angled Triangle, Trigonometric Ratios, Plane Geometry, Solid Geometry
Trigonometry
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Trigonometry deals with triangles, their properties, and theorems related to triangles. Trigonometry is often considered to be a part or subset of Geometry. While they are interrelated, trigonometry only focuses on right-angled triangles.
- Trigonometric theorems and formulas are also specific to right-angled triangles.
- It deals with trigonometric ratios used for the relationship between angles and the ratios of the side of a right-angled triangle.
- These ratios are named as sine, cosine, tangent, secant, cosecant, and cotangent.
Trigonometry studies the three sides of a right-angled triangle: hypotenuse, base, and perpendicular. The same is illustrated below.
Read Also: Introduction to Trigonometry Formulas
Geometry
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Geometry deals with almost all kinds of figures such as squares, rectangles, triangles, and also circles. Geometry is not only limited to two-dimensional shapes like squares and rectangles but also deals with three-dimensional shapes like cubes and cuboids. The geometry that deals with two-dimensional shapes is called Plane Geometry and the one that deals with three-dimensional shapes is called Solid Geometry.
Geometry
Read More:
| Trigonometry and Geometry Related Concepts | ||
|---|---|---|
| Perimeter of Shapes | 3D Shapes | Trigonometry Values |
| Trigonometric Functions | Trigonometric Identities | Surface Area of Cuboid |
| Similarities of Triangles | Areas Related to Circles | Surface Areas and Volumes |
Difference Between Trigonometry and Geometry
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The differences between trigonometry and geometry are as follows:
| Trigonometry | Geometry |
|---|---|
| Trigonometry only deals with the properties and features of triangles, specifically, right-angled triangles. | Geometry deals with properties and measurement of different types of shapes. |
| Trigonometry is part of geometry. | Geometry is not a subset of trigonometry rather it is its superset. |
| It deals with the relationship that exists between angles and sides of a triangle. | It deals with the different properties of the geometrical figures. |
| Hipparchus is considered to be the father/founder of Trigonometry. | Euclid is considered to be the father/founder of Geometry. |
Also Read: Isosceles Right-Angled Triangle
Important Formulas and Concepts
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The important formulas and concepts related to trigonometry and geometry are provided below.
Trigonometry
The most important concept in Trigonometry is the Pythagoras theorem as it forms the base for all trigonometric formulas. The theorem is as follows:
| Pythagoras theorem states that in a right-angled triangle, the square of hypotenuse is equal to the sum of squares of perpendicular and the base. |
Consider a triangle ABC,
where AB = P (Perpendicular), BC = B (Base), and CA = H (Hypotenuse)
Here, H2 = P2 + B2
The trigonometric ratios based on the theorem are:
| Sin A = P/H | Cot A = B/P |
| Cos A = B/H | Cosec A = H/P |
| Tan A = P/B | Sec A = H/B |
Here,
- tan A = sin A/cos A
- cot A = cos A/sin A
- cosec A = 1/sin A
- sec A = 1/cos A
Based on these ratios, some important formulas are given below:
- sin (90o-A) = Cos A
- cos (90O-A) = Sin A
- tan (90o-A) = cot A
- cot (90o-A) = tan A
- sec (90o-A) = cosec A
- sin2A = 2sinAcosA
- sin2A + cos2A = 1
- tan2A + 1 = sec2A
- tan2A = 2tanA/1 – tan2A
- cot2A= cot2A – 1/2cotA
Solved ExamplesQues. In a right-angled triangle, the base is 6 cm, and the perpendicular is 8 cm. Find the hypotenuse and also the values for sin, cos, and cot in the triangle. Sol. We will first find out the hypotenuse. H2 = P2 + B2 100 = 64 + 36 Thus, we get H = 10 cm Now, sin = P/H = 8/10 Cos = B/H = 6/10 and Cot = B/P = 6/8 Hence solved. Ques. In a right-angled triangle ABC, perpendicular AB is 24 cm and base BC is 7 cm. Find sin A and cos A. Sol. Using Pythagoras theorem, H2 = P2 + B2 (AC)2 = (24)2 + (7)2 (AC)2 = 576 + 49 (AC)2 = 625 cm2 AC = 25 cm Now, sin A = BC/AC = 7/25 cos A = AB/AC = 24/25 |
Geometry
Geometry formulas are used to measure area, surface area, and perimeter of a figure. Some of these formulas are given below:
- Area of Triangle = ½ x base x height
- Area of Square = side2
- Area of circle = π r2
- Area of Rectangle = length x breadth
- Perimeter of Rectangle = 2(length + breadth)
- Perimeter of Square = 4 x side
- Circumference of Circle = 2πr
Solved ExampleQues. Find the area and circumference of a circle with a diameter of 15 cm? Sol. Area of circle = 2πr And the circumference of a circle = πr2 Here, we first need to find the radius. R = d/2 Hence, r = 15/2 = 7.5 cm Now, Area = 22/7 x 7.5 x 7.5 = 176.78 cm2 Circumference = 2 x 22/7 x 7.5 = 47.14 cm |
Things to Remember
- Trigonometry is a subset of Geometry and deals with properties and ratios of right-angled triangles.
- Geometry deals with different geometrical figures such as triangles, squares, rectangles etc.
- Geometry that deals with two-dimensional figures is called plane geometry and the one which deals with three-dimensional figures is called solid geometry.
- The Pythagoras theorem is the base of trigonometry and all ratios and formulas are based on the theorem.
- Pythagoras theorem states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of squares of other two sides.
- Trigonometry deals with the ratios between angles and sides of right-angled triangles, which are called trigonometric ratios.
Sample Questions
Ques. A dimensions are as follows: l = 1.6 cm, b = 2.4cm, h= 2.6cm. Find the total surface area. (3 Marks)
Sol. The total surface area of a cuboid = 2 (lb + bh + hl)
Hence, the total surface area will be = 2 (1.6 × 2.4 + 2.4 × 2.6 + 2.6 × 1.6)
= 2 (3.84 + 6.24 + 4.2)
= 2 (14.28) = 28.56 cm2
Therefore, the total surface area of the cuboid will be 28.56 cm2.
Ques. If sin θ + cos θ = √3, then prove that tan θ + cot θ = 1. (CBSE Sample Question Paper 2019-20) (3 Marks)
Sol. sin θ + cos θ = √3
In the next step, we square both sides.
(sin θ + cos θ)2 = (√3)2
We get,
(sin θ + cos θ)2 = 3
Using identity,
Sin2 θ + cos2 θ + 2cos θ sin θ = 3
Using sin2 θ + cos2 θ = 1
1 + 2 cos θ sin θ = 3
Giving us,
cos θ sin θ = 1 …. (1)
Now to prove, tan θ + cot θ = 1
We will solve LHS here,
tan θ + cot θ = sin θ/ cos θ + cos θ/ sin θ
= sin2 θ + cos2 θ / cos θ sin θ
Here now we put sin2 θ + cos2 θ = 1
= 1 / cos θ sin θ
(And from (1) we get)
= 1/ 1
=1 = RHS
LHS = RHS
Hence proved.
Ques. An airplane’s angle of elevation from a point on the ground is 60o. After around 30 seconds, the angle of elevation becomes 30o. If the airplane is flying at a constant height of 3000√3m, find the speed of the airplane (CBSE Sample Question Paper 2019-20) (5 Marks)
Sol. Firstly, we name this point on the ground to be A.
Since the airplane is flying at around 3000√3 m above,
CD = BE = 3000√3 m
Now, finding the speed
distance travelled = BC
Assuming BC = x
∠ABE = 90O and ∠ACD = 900
Finding AB,
tan A= BE / AB
tan 600 = BE / AB
√3 = 3000 √3 / AB
AB = 3000 m
tan A = CD / AC
tan 300 = CD/ AC
1/ √3 = 3000√3/ AC
AC = 3000√3 x √3 = 9000 m
Here, AC = AB + BC
9000 = 3000 + BC
9000 – 3000 = BC
BC = 6000 m
Therefore, the distance travelled by plane = 6000 m.
Now since speed = distance/time, the speed of the plane will be
= 6000/ 30 m/s
= 200 m/s
Speed of the airplane = 200 m/s
Ques. A man in a boat is rowing away from a lighthouse 100 m high takes 2 minutes to change the angle of elevation of the top of the lighthouse from 60o to 30o. Find the speed of the boat (meters per minute). (CBSE 2019 Exam) (5 Marks)
Sol.
In the illustrated figure, CD = distance travelled by man which changes the angle of elevation from 30o to 60o
Distance = Speed x Time
Tan (60o) = AB / BC
√3 = 100 / BC
BC = 100 / √3
tan (30o) = AB / BD
1 / √3 = 100/ BD
BD = 100√3
CD = BD -BC
2x = 100√3 – 100/√3
2x = 300- 100/ √3
x = 200/ 2√3
x = 100/ √3
(√3 = 1.73)
x = 100/ 1.73 = 57.80
Therefore, we get the speed of the boat to be 57.80 m per minute.
Ques. Prove that (1 + tan A – Sec A) x (1 + tan A + Sec A) = 2 tan A (CBSE 2020 EXAM) (5 Marks)
Sol. We need to prove here LHS = RHS
We begin with LHS,
(1 + tan A – sec A) x (1 + tan A + sec A)
= (1 + tan A)2 – (sec A)2 {Using identity = [ (a+b)(a-b) = a2 – b2)]}
= 1 + tan2A + 2 tan A – sec2 A
= sec2 A + 2tan A – sec2 A
Putting, sec2 θ = 1 + tan2 θ
Gives us,
= 2 tan A
Hence, LHS = RHS
Hence proved.
Ques. Prove that (sin θ + cosec θ)2 + (cos θ + sec θ)2 = 7 + tan2 θ + cot2 θ (CBSE 2020 Exam) (5 Marks)
Sol. We need to prove here that LHS = RHS
We begin with LHS,
= (sin2 θ + cosec2 θ)2 + (cos θ + sec θ)2
= sin2 θ + cosec2 θ + 2 sin θ cosec θ + cos2 θ+ sec2 θ + 2 cos θ sec θ
= (sin2 θ + cos2 θ) + cosec2 θ + 2 sin θ x 1/ sin θ
+ sec2 θ + 2 cos θ x 1/ cos θ
= 1 + (1 + cot2 θ) + 2 + (1 + tan2 θ) + 2
= 7 + tan2 θ + cot2 θ
Hence, LHS = RHS
Ques. Prove that a Parallelogram that circumscribes a circle is also a rhombus. (5 Marks) (2019 CBSE)
Sol. Given: A circle with centre O and a Parallelogram ABCD touches the circle at points P, Q, R, and S.
Prove: ABCD is a Rhombus.
Proof: Since a rhombus is a parallelogram with equal sides. Here we need to prove that all sides of the parallelogram are equal.
The opposite sides of a parallelogram are equal.
Hence, AB = CD and AD = BC
According to a theorem, lengths of tangents that are drawn from an external point are always equal.
Hence,
AP = AS – (1), BP = BA – (2), CR = CQ - (3), DR = DS – (4)
AP + BP + CR + DR = AS + BQ + CQ + DS
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
AB + CD = AD + BC
(Since, CD = AB and BC = AD)
AB + AB = AD + AD
2AB = 2AD
AB = AD
And hence,
AB = AD
Further, AB = CD and AD =BC
Giving us, AB = CD = AD = CD
Hence, proving ABCD parallelogram has all sides equal.
Therefore, ABCD is a rhombus.
Ques. Prove that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides. (4 Marks) (2020 CBSE)
Sol. Given, ABC ~ PQR
Prove: (ar (ABC)) / (ar (PQR)) = (AB/PQ)2 = (BC/QR)2 = (AC/PR)2
Firstly, construct AM ⊥ BC and PN ⊥ QR
Proof: Area of a = ½ x base x height
Hence, ar (ABC) = ½ x BC x AM
Ar (PQR) = ½ x QR x PN
In ABM and PQN,
We find, ∠B = ∠Q,
∠M = ∠N,
Giving us,
ABM ~ PQN
Therefore, AB|PN = AM|PN
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