BalbhartiMaharashtra State Board Class 11 Math SolutionsPdf Chapter 2 Trigonometry – I Ex 2.2 Questions and Answers.

## Maharashtra State Board 11. Mathematical Solutions Chapter 2 Trigonometry – I Ex 2.2

Question 1.

Se A = 1 = \(\sqrt{2}\) cos B e \(\frac{\pi}{2}\) < A < π, \(\frac{3 \pi}{2}\)

Solution:

Given, 2sin A = 1

∴ you A = 1/2

We know,

why^{2}A = 1 - Ohne^{2}A = 1 – \(\left(\frac{1}{2}\right)^{2}=1-\frac{1}{4}=\frac{3}{4}\)

∴ cos A = \(\pm \frac{\sqrt{3}}{2}\)

Given that \(\frac{\pi}{2}\) < A < π

A is in the second quadrant.

We know,

sin^{2}B = 1 - because^{2}B = 1 – \(\left(\frac{1}{\square{2}}\right)^{2}\)\(\frac{1}{2}=\frac{1}{2}\ )

∴ sen B = \(\pm \frac{1}{\sqrt{2}}\)

Like \(\frac{3 \pi}{2}\) < B < 2π

B is in the fourth quadrant,

question 2

If \(\) and A, B are angles in the second quadrant, then prove that 4cosA + 3cosB = -5

Solution:

Sim \(\frac{\sin \mathrm{A}}{3}=\frac{\sin \mathrm{B}}{4}=\frac{1}{5}\)

∴ sen A = \(\frac{3}{5}\) y sen B = \(\frac{4}{5}\)

We know,

why^{2}A = 1 - Ohne^{2}= 1 – \(\left(\frac{3}{5}\right)^{2}\) = 1 – \(\frac{9}{25}=\frac{16}{25}\)

∴Cos A = ± \([{4}{5}\)

Since A is in the second quadrant,

because A < 0

∴ Because A = –\(\frac{4}{5}\)

Sine B = 4/5

We know,

why^{2}B = 1 - sin^{2}B = 1 – \(\left(\frac{4}{5}\right)^{2}=1-\frac{16}{25}=\frac{9}{25}\)

∴ Cos B = ±\(\frac{4}{5}\)

Since B is in the second quadrant, cos B < 0

question 3

If tan θ = \(\frac{1}{2}\), then \(\frac{2\sin\theta+3\cos\theta}{4\cos\theta+3\sin\theta}\)

Solution:

question 4

Remove 0 from the following:

I. x = 3secθ, y = 4tanθ

ii. x = 6 coseg θ, y = 8 coseg θ

iii. x = 4 cos θ – 5 sen θ, y = 4 sen θ + 5 cos θ

4. x = 5 + 6 cosec θ,y = 3 + 8 cot θ

v. x = 3 – 4. θ,3y = 5 + 3 Sek. θ

Solution:

I. x = 3secθ, y = 4tanθ

∴ sec θ = \(\frac{x}{3}\) y tan θ= \(\frac{y}{4}\)

We know,

sek^{2}θ - light brown^{2}θ = 1

∴ 16x^{2}- nine years old^{2}= 144

ii. x = 6 coseg θ y y = 8 cot θ

.'. cosec θ = \(\) y cot θ = \(\)

We know,

cosec^{2}θ—cuna^{2}θ =

16x^{2}- nine years old^{2}= 576

iii. x = 4cos θ – 5 sen θ ... (i)

y = 4sinθ + 5cosθ .. .(ii)

If we square (i) and (ii) and add, we get

x^{2}+j^{2}= (4cosθ – 5sinθ)^{2}+ (4 sen θ + 5 cos θ)^{2}

= 16 because^{2}θ – 40 senθ cosθ + 25 sen^{2}θ + 16 sin^{2}θ + 40 sen θ cos θ + 25 cos^{2}EU

= 16 (sin^{2}θ + cos^{2}i) + 25(sin^{2}θ + cos^{2}EU)

= 16(1) + 25(1)

= 41

4. x = 5 + 6 cosec θ undy = 3 + 8cot θ

∴ x – 5 = 6 coseg θ y y – 3 = 8 coseg θ

∴ cosec θ = \(\frac{x-5}{6}\) y cot θ = \(\frac{y-3}{8}\)

We know,

cosec^{2}θ—cuna^{2}θ = 1

∴ \(\left(\frac{x-5}{6}\right)^{2}-\left(\frac{y-3}{8}\right)^{2}\) = 1

v. 2x = 3 - 4then θ and 3y = 5 + 3sec θ

∴ 2x – 3 = -4tan θ y 3y – 5 = 3seg θ

∴ tan θ = \(\frac{3-2 x}{4}\) y seg θ = \(\frac{3 y-5}{3}\)θ

We know that, sec.^{2}θ - light brown^{2}θ = 1

∴ \(\left(\frac{3 y-5}{3}\right)^{2}-\left(\frac{3-2 x}{4}\right)^{2}\) = 1

∴ \(\left(\frac{3 y-5}{3}\right)^{2}-\left(\frac{2 x-3}{4}\right)^{2}\) = 1

question 5

say 2 sen^{2}θ + 3sin θ = 0, find the allowed values of cosθ.

Solution:

2 sin^{2}θ + 3sen θ = 0

∴ sin θ (2 sin θ + 3) = 0

∴ sen θ = 0 o sen θ = \(\frac{-3}{2}\)

Como - 1 ≤ sin θ ≤ 1,

sen θ = 0

\(\sqrt{1-\cos^{2}\theta}\) = 0 ...[ ∵ Som^{2}θ = 1- pig^{2}EU]

∴ 1 – because^{2}θ = 0

∴ because^{2}θ = 1

∴ cos θ = ±1 …[∵ – 1 ≤ cos θ ≤ 1]

question 6

yes 2 because^{2}θ – 11 cos θ + 5 = 0, then find the possible values of cos θ.

Solution:

2 because^{2}θ – 11 cos θ + 5 = 0

∴ 2cos^{2}θ – 10 cos θ – cos θ + 5 = 0

∴ 2cos θ(cos θ – 5) – 1 (cos θ – 5) = 0

∴ (cos θ – 5) (2cos θ – 1) = 0

cos θ – 5 = 0 or 2 cos θ – 1 = 0

∴ cos θ = 5 o cos θ = 1/2

Da -1 ≤ cos θ ≤ 1

∴ cos θ = 1/2

question 7

Find the acute angle θ as 2cos^{2}θ = 3sen θ.

Solution:

2 because^{2}0 = 3 senθ

∴ 2(1 – sin^{2}i) = 3sen i

∴2–2 sen^{2}θ = 3sen θ

∴2sünde^{2}θ + 3sen 9-2 = θ

∴2sünde^{2}θ + 4sen θ – sen θ – 2 = θ

∴ 2sen θ(sen θ + 2) -1 (sen θ + 2) = θ

∴ (sen θ + 2) (2sen θ – 1) = 0

∴ sen θ + 2 = 0 o 2 sen θ – 1 = 0

∴ sen θ = -2 o sen θ = 1/2

Like, -1 ≤ sin θ ≤ 1

∴ Sen θ = 1/2

∴ θ = 30° …[ ∵ sen 30 = 1/2]

question 8

Find the acute angle 0 such that 5tan2 0 + 3 = 9sec 0.

Solution:

5 tan^{2}θ + 3 = 9 seconds θ

∴ 5(sec^{2}θ – 1) + 3 = 9 seconds. θ

∴ 5 seconds^{2}θ – 5 + 3 = 9 seconds. θ

∴ 5 seconds^{2}θ – 9 seconds. θ – 2 = 0

∴ 5 seconds^{2}i – 10 seconds. θ + sec. θ – 2 = 0

∴ 5 sec. θ(Seg. θ – 2) + 1(Seg. θ – 2) = 0

∴ (Seg. θ – 2) (5 Seg. θ + 1) = 0

∴ seg θ – 2 = 0 o 5 seg θ + 1 = 0

∴ second θ = 2 the second θ = -1/5

Given that sec θ ≥ 1 or sec θ ≤ -1,

Second. θ = 2

∴ θ = 60° … [ ∵ Seg. 60° = 2]

question 9

Encontre sen θ tal que 3cos θ + 4sin θ = 4.

Solution:

3cosθ + 4sinθ = 4

∴ 3cos θ = 4(1 – Sünde θ)

If we square both sides, we get .

9 because^{2}θ = 16(1 – Sunde θ)^{2}

∴ 9(1 – sin^{2}θ) = 16(1 + sen^{2}θ – 2sen θ)

∴9–9 domingo^{2}θ = 16 + 16 sen^{2}me - 32sen me

∴25sünde^{2}θ – 32sen θ + 7 = 0

∴25sünde^{2}θ – 25 sen θ – 7 sen θ + 7 = 0

25sen θ (sen θ – 1) – 7 (sen θ – 1) = 0

∴ (sen θ – 1) (25sen θ – 7) = 0

∴ sen θ – 1 = 0 o 25 sen θ – 7 = 0

∴ sen θ = 1 o sen θ = \(\frac{7}{25}\)

Like, -1 ≤ sin θ ≤ 1

∴ sen θ = 1 o \(\frac{7}{25}\)

[Hint: The answer given in the book is 1. However, by our calculation it is 1 or \(\frac{7}{25}\).]

question 10

Se cosec θ + cot θ = 5, then calculate sec θ

Solution:

cosec θ + berço θ = 5

∴ \(\frac{1}{\sin \theta}+\frac{\cos \theta}{\sin \theta}=5\)

∴ \(\frac{1+\cos\theta}{\sin\theta}=5\)

∴ 1 + cosθ = 5.senθ

If we square both sides, we get

1 + 2 cos θ + cos^{2}θ = 25sen^{2}EU

∴ because^{2}θ + 2 cos θ + 1 = 25 (1 – cos^{2}EU)

∴ because^{2}θ + 2 cos θ + 1 = 25 – 25 cos^{2}EU

∴ 26 because^{2}θ + 2 because θ – 24 = 0

∴ 26 because^{2}θ + 26 cos θ – 24 cos θ – 24 = 0

∴ 26 cos θ (cos θ + 1) – 24 (cos θ + 1) = 0

∴ (cos θ + 1) (26 cos θ – 24) = 0

∴ cos θ + 1 = θ o 26 cos θ – 24 = 0

∴ cos θ = -1 o cos θ = \(\frac{24}{26}=\frac{12}{13}\)

Wenn cos θ = -1, sin θ = 0

∴ cot θ and cosec x are not defined,

∴ cosθ ≠ -1

∴ cos θ = \(\frac{12}{13}\)

∴ seg. θ = \(\frac{1}{\cos \theta}=\frac{13}{12}\)

[Note: the answer given in the book is -1 or \(\frac{13}{12}\).

However, by our calculations, it's just \(\frac{13}{12}\).]

question 11

If θ = \(\frac{3}{4}\) e π < θ < \(\frac{3 \pi}{2}\), I find a value of 4 coseg θ + 5 cos θ.

Solution:

We know,

cosec^{2}θ = 1 + cuna^{2}θ = \(\left(\frac{3}{4}\right)^{2}\) = 1 + \(\frac{9}{16}\)

∴ cosek^{2}θ = \(\frac{25}{16}\)

∴ cosec θ = \(\pm \frac{5}{4}\)

Da π < θ < \(\frac{3 \pi}{2}\)

θ is in the third quadrant.

∴ coseg θ < 0

∴ coseg θ = –\(\frac{5}{4}\)

Cuna θ = \(\frac{3}{4}\)

tan θ = \(\frac{1}{\cot \theta}=\frac{4}{3}\)

We know,

sek^{2}θ = 1 + tan^{2}θ = 1 + \(\left(\frac{4}{3}\right)^{2}\)

= 1 + \(\frac{16}{9}=\frac{25}{9}\)

∴ second θ = ±\(\frac{5}{3}\)

Since θ is in the third quadrant,

Second. θ < 0

∴ second θ = –\(\frac{5}{3}\)

cos θ = \(\frac{1}{\sec \theta}=\frac{-3}{5}\)

∴ 4 coseg θ + 5 cos θ

= \(4\left(-\frac{5}{4}\right)+5\left(-\frac{3}{5}\right)\)

= -5 – 3 = -8

[Note: the question has been modified.]

question 12

Find the Cartesian coordinates of the points whose polar coordinates are:

I. (3, 90°) ii. (1, 180 degrees)

Solution:

I. (r, θ) = (3, 90°)

Using x = r cos θ and y = r sin θ, where (x,y) are the necessary Cartesian coordinates, we get

x = 3cos 90° y y = 3sen 90°

∴ x = 3(0) = 0 e y = 3(1) = 3

∴ the necessary Cartesian coordinates are (0, 3).

ii. (r, θ) = (1, 180°)

Using x = r cos θ and y = r sin θ, where (x,y) are the necessary Cartesian coordinates, we get

x = 1(cos 180°) e y = 1(sen 180°)

∴ x = -1 e y = 0

∴ the necessary Cartesian coordinates are (-1, 0).

question 13

Find the polar coordinates of the points whose Cartesian coordinates are:

1. (5, 5) ii. (1, \(\square root{3}\))

ii. (-1, -1) iv. (-\(\square root{3}\), 1)

Solution:

I. (x, y) = (5, 5)

∴ r = \(\square root{x^{2}+y^{2}}\) = \(\square root{25+25}\)

\(=\square root{50}=5 \square root{2}\)

tan θ = \(\frac{y}{x}=\frac{5}{5}\) = 1

Since the given point is in the 1st quadrant,

θ = 45° …[∵ tan 45° = 1]

∴ the required polar coordinates are (\(5 \sqrt{2}\), 45°).

ii. (x, y) = (1, \(\sqrt{3}\))

∴ r = \(\square root{x^{2}+y^{2}}=\square root{1+3}=\square root{4}=2\)

tan θ = \(\frac{y}{x}=\frac{\sqrt{3}}{1}=\sqrt{3}\)

Since the given point is in the 1st quadrant,

θ = 60° …[∵ tan 60° = \(\sqrt{3}\)]

∴ the required polar coordinates are (2, 60°).

iii. (x, y) = (-1, -1)

∴ r = \(\square root{x^{2}+y^{2}}=\square root{1+1}=\square root{2}\)

tan θ = \(\frac{y}{x}=\frac{-1}{-1}=1\)

∴ tan θ = tan \(\frac{\pi}{4}\)

Since the given point is in the third quadrant,

tan θ = tan \(\left(\pi+\frac{\pi}{4}\right)\) ...[∵ tan(n + x) = tanx]

∴ tan θ = tan \(\left(\frac{5\pi}{4}\right)\)

∴ θ = \(\frac{5 \pi}{4}\) = 225°

∴ the required polar coordinates are (\(\sqrt{2}\), 225°).

4. (x, y) = (-\(\sqrt{3}\) , 1)

∴ r = \(\square root{x^{2}+y^{2}}=\square root{3+1}=\square root{4}=2\)

tan θ = \(\frac{y}{x}=\frac{1}{-\sqrt{3}}\) = -tan \(\frac{\pi}{6}\)

Since the specified point is in the second quadrant,

tan θ = tan \(\left(\pi-\frac{\pi}{6}\right)\) ...[∵ tan (π – x) = – tanx]

∴ tan θ = tan \(\left(\frac{5\pi}{6}\right)\)

∴ θ = \(\frac{5 \pi}{6}\) = 150°

∴ the required polar coordinates are (2, 150°)

question 14

Find the values of:

I. Sin\(\frac{19 \pi^{e}}{3}\)

ii. Korb 1140°

iii. Cuna \(\frac{25 \pi^{e}}{3}\)

Solution:

I. We know that the sine function is periodic with period 2π.

sin \(\frac{19 \pi}{3}\) = sin \(\left(6 \pi+\frac{\pi}{3}\right)\) = sin \(\frac{\pi}{ 3}=\frac{\sqrt{3}}{2}\)

ii. We know that the cosine function is periodic with period 2π.

cos 1140° = cos (3 × 360° + 60°)

= cos 60° = \(\frac{1}{2}\)

iii. We know that the cotangent function is periodic with period π.

Children's bed \(\frac{25 \pi}{3}\) = Children's bed \(\left(8 \pi+\frac{\pi}{3}\right)\) = Children's bed \(\frac{\ pi}{ 3}\) = \(\frac{1}{\sqrt{3}}\)

dhana work.txt

See the work of dhana.txt.

## FAQs

### Is Class 11 trigonometry hard? ›

In Class 11 **Trigonometric Functions is the easiest one among all other chapters**. Though there are a number of formulas, students can easily score the marks for this chapter.

**Which chapter is trigonometry in class 11? ›**

NCERT Solutions Class 11 Maths **Chapter 3** Trigonometric Functions- 2022-23 Free PDF Download.

**Is there trigonometry in class 11 Maths? ›**

So basically, trigonometry is a study of triangles, which has angles and lengths on its side. Trigonometry basics consist of sine, cosine and tangent functions. **Trigonometry for class 11 contains trigonometric functions**, identities to solve complex problems more simply.

**How many chapters are there in Maths class 11 Maharashtra board? ›**

Answer: The current 11th Maths and Statistics syllabus Maharashtra board includes **18 chapters** which are divided into two parts.

**Is Class 11 hard or 12? ›**

Hi **11th standard is comparatively tougher than 12th**. so before starting 11th class go through the basics and formulas related to 11th class and try to solve questions of math, physics, chemistry from NCERT book.

**Is trigonometry easy or hard? ›**

**Trigonometry is considered hard** by students due to the following reasons: The memorization of formulas and values. Non-linear functions. Angle measurement in Radians/Degree.

**What is the hardest chapter in class 11 Maths? ›**

**Circle, Parabola and Permutation and Combination** are tough chapters of Class 11 Maths. Sequence and Series is another tough topic that needs more attention and preparation. Additionally, you must also study Coordinate Geometry and Integral Calculus for JEE Mains 2022.

**Which chapters are important in Maths 11? ›**

**Important Class 11 Maths Topics from Each Chapter**

- Cartesian Product of Sets.
- Concept of Relations.
- Real-Valued Functions.

**Is trigonometry an important chapter? ›**

**Trigonometry is important to mathematics** as it involves the study of calculus, statistics and linear algebra. This article covers the identities of trigonometry and trigonometric equations, functions of trigonometry, properties of inverse trigonometric functions, problems on heights and distances.

**Can we skip Maths in Class 11? ›**

To pursue a career in these fields, you are not required to be a genius in Mathematics. However, a good hold over numbers is surely required. Knowledge of statistics and analytical skills are a must, and thus **maths till class 12 ^{th} can be beneficial, even if it is not compulsory.**

### How do you solve trigonometry questions easily? ›

**Make a point of memorizing them.**

- Quotient Identities: tan(x) = sin(x)/cos(x) cot(x) = cos(x)/sin(x)
- Reciprocal Identities: csc(x) = 1/sin(x) sec(x) = 1/cos(x) cot(x) = 1/tan(x) sin(x) = 1/csc(x) ...
- Pythagorean Identities: sin
^{2}(x) + cos^{2}(x) = 1. cot^{2}A +1 = csc^{2}A. 1+tan^{2}A = sec^{2}A.

**What grade math is trigonometry? ›**

In general, trigonometry is taken as part of **sophomore or junior year math**. In addition to being offered as its own course, trigonometry is often incorporated as a unit or semester focus in other math courses.

**Which chapter has more weightage in Maths class 11? ›**

**Algebra** is the highest weightage unit of the CBSE 11th Maths syllabus which is followed by Sets and Function with the weightage of 23 marks.

**Which chapters are removed from class 11? ›**

CBSE Class 11 History Deleted Syllabus 2023: FAQs

The chapters which have been removed from Class 11 history syllabus are **Early Societies, Nomadic Empires, and Confrontation of Cultures**. Will syllabus be reduced for Class 11? Yes syllabus has been reduced form Class 11 history for the academic year 2021-2022.

**Which topics are removed from Maths 11? ›**

Maths Class 11 Deleted Syllabus 2022-23

**Difference of sets**. General Solutions of trigonometric equations of the type sin y = sin a, cos y = cos a, and tan y = tan a. Polar representation of complex numbers. Square root of a complex number.

**Why is 11th so hard? ›**

11^{th} is very different in terms of the course content as compared to what you have studied till 10^{th}. **Difficulty level is higher if you have chosen the science stream as now it is no more general science**. 11^{th} grade will teach you concepts of fundamental physics, chemistry, biology, maths etc.

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Totally depends on your school/college. They might take a re-test or just promote you directly to 12th. There is nothing to worry about. **They never let you fail you in class 11th**.

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Although the basic criteria to pass class 11 in each subjects of the science stream is about to **score at least 21 in all of the subjects out of 70**, as per the state board marking rules. However in other boards as well in class 11 you need to score at least 30% marks in each and every subjects.

**How can I pass trigonometry? ›**

**How to Pass Trigonometry**

- Review What You Have Learned in Previous Classes. Before beginning your trig class, make sure you start off on solid footing. ...
- Be a Star Student. Strive to attend every class. ...
- Trig Doesn't End at the Classroom Door. ...
- Ask for Help. ...
- Making Time. ...
- Keep Your Memory Sharp!

**What's the hardest math? ›**

**The Riemann Hypothesis**, famously called the holy grail of mathematics, is considered to be one of the toughest problems in all of mathematics.

### How do you study trigonometry fast? ›

**What is the Easiest Way to Learn Trigonometry?**

- Study all the basics of trigonometric angles.
- Study right-angle triangle concepts.
- Pythagoras theorem.
- Sine rule and Cosine rule.
- List all the important identities of trigonometry.
- Remember the trigonometry table.
- Be thorough with the trigonometric formulas.

**Which is the toughest chapter in maths class 11? ›**

**Circle, Parabola and Permutation and Combination** are tough chapters of Class 11 Maths. Sequence and Series is another tough topic that needs more attention and preparation. Additionally, you must also study Coordinate Geometry and Integral Calculus for JEE Mains 2022.

**Which is difficult subject in 11th? ›**

**Science (PCM or PCB)**

Science stream has a reputation of most difficult stream in class 11th and 12th. Students learn Science till 10th standard, so they are aware of the basics of this subject.

**Is trigonometry a hard class? ›**

**Trigonometry is hard because it deliberately makes difficult what is at heart easy**. We know trig is about right triangles, and right triangles are about the Pythagorean Theorem. About the simplest math we can write is When this is the Pythagorean Theorem, we're referring to a right isosceles triangle.

**Is trigonometry harder than precalculus? ›**

Consequently, some students have a steep learning curve upon entering precal and will feel like they are swimming in unchartered waters for a while. Now, most students agree that **math analysis is “easier” than trigonometry**, simply because it's familiar (i.e., it's very similar to algebra).

**Why is 11th standard hard? ›**

11^{th} is very different in terms of the course content as compared to what you have studied till 10^{th}. **Difficulty level is higher if you have chosen the science stream as now it is no more general science**. 11^{th} grade will teach you concepts of fundamental physics, chemistry, biology, maths etc.

**Which is the easiest chapter in maths class 11? ›**

Generally easiness of chapter in Mathematics depends upon the priority of a person's interest. However some chapters like **set theory, mathematical induction, A.P, G.P, differentiation, matrix** in class 11 and 12 are considered to less complicated compared to other chapters.

**Which chapter is most important in maths class 11? ›**

**Important Class 11 Maths Topics from Each Chapter**

- Cartesian Product of Sets.
- Concept of Relations.
- Real-Valued Functions.

**What is the most failed course in high school? ›**

**Algebra** is the single most failed course in high school, the most failed course in community college, and, along with English language for nonnative speakers, the single biggest academic reason that community colleges have a high dropout rate.

**Which is the most difficult grade in school? ›**

While **junior year** is often the hardest year of high school, the transition from middle school to 9th grade can also be tough. To make it easier, don't feel afraid to reach out to your teachers and counselors, and take advantage of the support resources that are available.

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**Which math class is hardest? ›**

What is the Hardest Math Class in High School? In most cases, you'll find that **AP Calculus BC or IB Math HL** is the most difficult math course your school offers. Note that AP Calculus BC covers the material in AP Calculus AB but also continues the curriculum, addressing more challenging and advanced concepts.

**What grade level is trigonometry? ›**

In general, trigonometry is taken as part of **sophomore or junior year math**. In addition to being offered as its own course, trigonometry is often incorporated as a unit or semester focus in other math courses.

**Is calculus or trigonometry easier? ›**

BTW **basic calculus is not easier than basic trigonometry** because basic trigonometric essentially deals with algebraic properties of circular functions. Calculus on the other hand is essentially non-algebraic in nature.

**Is trigonometry the most important math? ›**

As **one of the most important branches of mathematics**, trigonometry is something that every student should focus on. Great trigonometry skills allow students to work out complex angles and dimensions in relatively little time.