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## Trigonometry Problems - sin, cos, tan, cot: Problems with Solutions

## Copy Previous "Geometry" Construction?

I would like to basically “Copy Previous” with a Geometry construction Desmos activity builder. I tried learn.desmos.com/copy-previous , but I don’t see 3 dots (I’m guessing because I have #CL enabled?) and I can’t figure out the sink (I think that’s the right word) for a Geometry component.

I want to do what this activity does from screen 5 to screen 6: Practical Trigonometry Problem Solving • Activity Builder by Desmos

If it’s helpful, this is my activity, and I want to copy the student’s construction on screen 5 onto screen 6: Triangle Congruence Theorems • Activity Builder by Desmos

Caught this one on twitter before I caught it here, my apologies. Let me know if you need help hooking up the geometry stuff in a graph.

Thanks, but I don’t think I can do what I want them to do outside of a Geometry component. The idea is to explore the triangle congruence theorems and show that the criteria meet our (transformation-based) definition of congruence. So they would need to create a triangle with, say, the same side lengths as a given triangle. Then, they would need to show that you can map the original triangle onto the new one using transformations. (And then wiggle the original triangle to be convinced that is always the case.)

They can do all that using a Geometry window, but it’s a lot of steps on one page. And I think that transferring it to a Graph window would make it complex enough that the students couldn’t actually do it themselves, which loses the constructed authenticity I’m aiming for. I’ll ponder whether I can get to a similar understanding in a different way.

I really want to use Desmos Geometry in my classes, but I guess it’s not there yet this year either. (I need CL in AB, and an Angle Bisector tool on desmos.com/geometry.) Thank you, though!!

Ooh yeah that’s tough. There are def ways to do it but probably not as clean looking as you’d like. Maybe take advantage of mathTHENexplain so get the answers you want.

I’ve taken to using buttons and the hidden sink to help navigate the large amount of expository text for long geometric constructions as my hack for keeping things on one screen.

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## Trigonometry Practice Questions

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## Free Mathematics Tutorials

Solve problems using trigonometric ratios.

A set of problems, that may be solved using the trigonometric ratios, is presented. Detailed solutions and explanations are included.
In a right triangle, the six trigonometric ratios; the ratio, the ratio, the ratio, the ratio, the ratio and the ratio are defined as follows: The sine of angle A = sin (A) = side opposite angle A / hypotenuse = a / c The cosine of angle A = cos (A) = side adjacent to angle A / hypotenuse = b / c The tangent of angle A = tan (A) = side opposite angle A / side adjacent to angle A = a / b The secant of angle A = sec (A) = hypotenuse / side adjacent to angle A = c / b The cosecant of angle A = csc (A) = hypotenuse / side opposite to angle A = c / a The cotangent of angle A = cot (A) = side adjacent to angle A / side opposite angle A = b / a
(hypotenuse) = 8 + 6 = 100 and hypotenuse = 10 sin A = side opposite angle A / hypotenuse = 8 / 10 = 4 / 5 cos (A) = side adjacent to angle A / hypotenuse = 6 / 10 = 3 / 5 tan (A) = side opposite angle A / side adjacent to angle A = 8 / 6 = 4 / 3 sec (A) = hypotenuse / side adjacent to angle A = 10 / 6 = 5 / 3 csc (A) = hypotenuse / side opposite to angle A = 10 / 8 = 5 / 4 cot (A) = side adjacent to angle A / side opposite angle A = 6 / 8 = 3 / 4 sin A = opposite / hypotenuse sin 31 = 5.12 / c c = 5.12 / sin 31 c (approximately) = 9.94 sin 30 = x / 10 = 1 / 2 ( see ) to find x x = 5 10 = 5 + y y = √(75) = 5 √ (3) |

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## Trigonometry Questions

Trigonometry questions given here involve finding the missing sides of a triangle with the help of trigonometric ratios and proving trigonometry identities. We know that trigonometry is one of the most important chapters of Class 10 Maths. Hence, solving these questions will help you to improve your problem-solving skills.

What is Trigonometry?

The word ‘trigonometry’ is derived from the Greek words ‘tri’ (meaning three), ‘gon’ (meaning sides) and ‘metron’ (meaning measure). Trigonometry is the study of relationships between the sides and angles of a triangle.

The basic trigonometric ratios are defined as follows.

sine of ∠A = sin A = Side opposite to ∠A/ Hypotenuse

cosine of ∠A = cos A = Side adjacent to ∠A/ Hypotenuse

tangent of ∠A = tan A = (Side opposite to ∠A)/ (Side adjacent to ∠A)

cosecant of ∠A = cosec A = 1/sin A = Hypotenuse/ Side opposite to ∠A

secant of ∠A = sec A = 1/cos A = Hypotenuse/ Side adjacent to ∠A

cotangent of ∠A = cot A = 1/tan A = (Side adjacent to ∠A)/ (Side opposite to ∠A)

Also, tan A = sin A/cos A

cot A = cos A/sin A

Also, read: Trigonometry

## Trigonometry Questions and Answers

1. From the given figure, find tan P – cot R.

From the given,

In the right triangle PQR, Q is right angle.

By Pythagoras theorem,

PR 2 = PQ 2 + QR 2

QR 2 = (13) 2 – (12) 2

= 169 – 144

tan P = QR/PQ = 5/12

cot R = QR/PQ = 5/12

So, tan P – cot R = (5/12) – (5/12) = 0

sin (90° – A) = cos A cos (90° – A) = sin A tan (90° – A) = cot A cot (90° – A) = tan A sec (90° – A) = cosec A cosec (90° – A) = sec A
cos A + sin A = 1 1 + tan A = sec A cot A + 1 = cosec A |

2. Prove that (sin 4 θ – cos 4 θ +1) cosec 2 θ = 2

L.H.S. = (sin 4 θ – cos 4 θ +1) cosec 2 θ

= [(sin 2 θ – cos 2 θ) (sin 2 θ + cos 2 θ) + 1] cosec 2 θ

Using the identity sin 2 A + cos 2 A = 1,

= (sin 2 θ – cos 2 θ + 1) cosec 2 θ

= [sin 2 θ – (1 – sin 2 θ) + 1] cosec 2 θ

= 2 sin 2 θ cosec 2 θ

= 2 sin 2 θ (1/sin 2 θ)

3. Prove that (√3 + 1) (3 – cot 30°) = tan 3 60° – 2 sin 60°.

LHS = (√3 + 1)(3 – cot 30°)

= (√3 + 1)(3 – √3)

= 3√3 – √3.√3 + 3 – √3

= 2√3 – 3 + 3

RHS = tan 3 60° – 2 sin 60°

= (√3) 3 – 2(√3/2)

= 3√3 – √3

Therefore, (√3 + 1) (3 – cot 30°) = tan 3 60° – 2 sin 60°.

Hence proved.

4. If tan(A + B) = √3 and tan(A – B) = 1/√3 ; 0° < A + B ≤ 90°; A > B, find A and B.

tan(A + B) = √3

tan(A + B) = tan 60°

A + B = 60°….(i)

tan(A – B) = 1/√3

tan(A – B) = tan 30°

A – B = 30°….(ii)

Adding (i) and (ii),

A + B + A – B = 60° + 30°

Substituting A = 45° in (i),

45° + B = 60°

B = 60° – 45° = 15°

Therefore, A = 45° and B = 15°.

5. If sin 3A = cos (A – 26°), where 3A is an acute angle, find the value of A.

sin 3A = cos(A – 26°); 3A is an acute angle

cos(90° – 3A) = cos(A – 26°) {since cos(90° – A) = sin A}

⇒ 90° – 3A = A – 26

⇒ 3A + A = 90° + 26°

⇒ 4A = 116°

⇒ A = 116°/4

6. If A, B and C are interior angles of a triangle ABC, show that sin (B + C/2) = cos A/2.

We know that, for a given triangle, the sum of all the interior angles of a triangle is equal to 180°

A + B + C = 180° ….(1)

B + C = 180° – A

Dividing both sides of this equation by 2, we get;

⇒ (B + C)/2 = (180° – A)/2

⇒ (B + C)/2 = 90° – A/2

Take sin on both sides,

sin (B + C)/2 = sin (90° – A/2)

⇒ sin (B + C)/2 = cos A/2 {since sin(90° – x) = cos x}

7. If tan θ + sec θ = l, prove that sec θ = (l 2 + 1)/2l.

tan θ + sec θ = l….(i)

We know that,

sec 2 θ – tan 2 θ = 1

(sec θ – tan θ)(sec θ + tan θ) = 1

(sec θ – tan θ) l = 1 {from (i)}

sec θ – tan θ = 1/l….(ii)

tan θ + sec θ + sec θ – tan θ = l + (1/l)

2 sec θ = (l 2 + 1)l

sec θ = (l 2 + 1)/2l

8. Prove that (cos A – sin A + 1)/ (cos A + sin A – 1) = cosec A + cot A, using the identity cosec 2 A = 1 + cot 2 A.

LHS = (cos A – sin A + 1)/ (cos A + sin A – 1)

Dividing the numerator and denominator by sin A, we get;

= (cot A – 1 + cosec A)/(cot A + 1 – cosec A)

Using the identity cosec 2 A = 1 + cot 2 A ⇒ cosec 2 A – cot 2 A = 1,

= [cot A – (cosec 2 A – cot 2 A) + cosec A]/ (cot A + 1 – cosec A)

= [(cosec A + cot A) – (cosec A – cot A)(cosec A + cot A)] / (cot A + 1 – cosec A)

= cosec A + cot A

9. Prove that: (cosec A – sin A)(sec A – cos A) = 1/(tan A + cot A)

[Hint: Simplify LHS and RHS separately]

LHS = (cosec A – sin A)(sec A – cos A)

= (cos 2 A/sin A) (sin 2 A/cos A)

= cos A sin A….(i)

RHS = 1/(tan A + cot A)

= (sin A cos A)/ (sin 2 A + cos 2 A)

= (sin A cos A)/1

= sin A cos A….(ii)

From (i) and (ii),

i.e. (cosec A – sin A)(sec A – cos A) = 1/(tan A + cot A)

10. If a sin θ + b cos θ = c, prove that a cosθ – b sinθ = √(a 2 + b 2 – c 2 ).

a sin θ + b cos θ = c

Squaring on both sides,

(a sin θ + b cos θ) 2 = c 2

a 2 sin 2 θ + b 2 cos 2 θ + 2ab sin θ cos θ = c 2

a 2 (1 – cos 2 θ) + b 2 (1 – sin 2 θ) + 2ab sin θ cos θ = c 2

a 2 – a 2 cos 2 θ + b 2 – b 2 sin 2 θ + 2ab sin θ cos θ = c 2

a 2 + b 2 – c 2 = a 2 cos 2 θ + b 2 sin 2 θ – 2ab sin θ cos θ

a 2 + b 2 – c 2 = (a cos θ – b sin θ ) 2

⇒ a cos θ – b sin θ = √(a 2 + b 2 – c 2 )

## Video Lesson on Trigonometry

## Practice Questions on Trigonometry

Solve the following trigonometry problems.

- Prove that (sin α + cos α) (tan α + cot α) = sec α + cosec α.
- If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
- If sin θ + cos θ = √3, prove that tan θ + cot θ = 1.
- Evaluate: 2 tan 2 45° + cos 2 30° – sin 2 60°
- Express cot 85° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.

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Armed with a clinometer and a tape measure, students are required to solve a series of trigonometry problems. What separates this activity from other trigonometry problems is that measurements aren't just laid out for students to plug into a formula, and so problems have an extra layer of choice. Text booky problems become non-text booky, and ...

Practical Trigonometry Problem Solving. Overview. Armed with a clinometer and a tape measure, students are required to solve a series of trigonometry problems. ... (Perhaps because they reveal answers or require a whole class conversation for introduction.)

Practical Trigonometry Problem Solving • Teacher Guide ... Loading...

Explore math with our beautiful, free online graphing calculator. Graph functions, plot points, visualize algebraic equations, add sliders, animate graphs, and more.

Trig Visual Calculator | Desmos. The following helps to visualize trigonometry. The circle on the right has a radius of 1 (indicated with the blue line). Each of the other colored lines represents one of the 6 fundamental trig functions. Here's what the colors stand for: Purple - Cosine; Red - Sine; Green - Tangent; Orange - Cotangent.

Prove the trigonometric identity: [tex]cos\alpha+cos2\alpha+cos6\alpha+cos7\alpha=4cos\alpha{\frac{\alpha}{2}}cos{\frac{5\alpha}{2}}cos4\alpha[/tex]

I want to do what this activity does from screen 5 to screen 6: Practical Trigonometry Problem Solving • Activity Builder by Desmos. If it's helpful, this is my activity, and I want to copy the student's construction on screen 5 onto screen 6: Triangle Congruence Theorems • Activity Builder by Desmos. Thank you!

Math 175: Plane Trigonometry. Chapter 3: Trigonometric Identities and Equations. Expand/collapse global location. 3.7: Exercises - Double Angle, Half-Angle, and Power Reductions. Page ID. Chapter 3 Practice: 1. If sin(x) = 1 8 and x is in quadrant I, then find exact values for (without solving for x): a. sin(2x)

Armed with a clinometer and a tape measure, students are required to solve a series of trigonometry problems. What separates this activity from other trigonometry problems is that measurements aren't just laid out for students to plug into a formula, and so problems have an extra layer of choice. Text booky problems become non-text booky, and the extra layer of choice can sometimes put your ...

Section 1.4 : Solving Trig Equations. Without using a calculator find the solution (s) to the following equations. If an interval is given find only those solutions that are in the interval. If no interval is given find all solutions to the equation. 4sin(3t) = 2 4 sin. . (3 t) = 2 Solution. 4sin(3t) = 2 4 sin. .

Explore math with our beautiful, free online graphing calculator. Graph functions, plot points, visualize algebraic equations, add sliders, animate graphs, and more.

Answers - Version 1. Answers - Version 2. The Corbettmaths Practice Questions on Trigonometry.

They do not store directly personal information, but are based on uniquely identifying your browser and internet device. If you do not allow these cookies, you will experience less targeted advertising. Free math problem solver answers your trigonometry homework questions with step-by-step explanations.

[Old version] Practical Trigonometry Problem Solving • Teacher Guide ... Loading...

Trigonometry (Honors) Review 3. Practice Questions (and Answers)Topics include trig values, half-angle identities, angular distance, quadrants a. Mathplane.com. Questions - Sol. tions - Thank. for visiting. (Hope it helped!)If you have questions, sugge. tions, Also, at Facebook, Google+, Pinterest, TES and TeachersPayTeachers.

Trigonometry problems with detailed solution are presented. Problem 1: A person 100 meters from the base of a tree, observes that the angle between the ground and the top of the tree is 18 degrees. Estimate the height h of the tree to the nearest tenth of a meter. Solution to Problem 1: Use the tangent. tan (18 o) = h / 100. Solve for h to obtain.

Some Right Angled Trig Practice from @JennSWhite https://teacher.desmos.com/activitybuilder/custom/5e7137882bccc050a066e045 Then some problem solving from ...

Solution to Problem 1: First we need to find the hypotenuse using Pythagora's theorem. (hypotenuse) 2 = 8 2 + 6 2 = 100. and hypotenuse = 10. We now use the definitions of the six trigonometric ratios given above to find sin A, cos A, tan A, sec A, csc A and cot A. sin A = side opposite angle A / hypotenuse = 8 / 10 = 4 / 5.

Explore math with our beautiful, free online graphing calculator. Graph functions, plot points, visualize algebraic equations, add sliders, animate graphs, and more.

Note that the point of these problems is not really to learn how to find the value of trig functions but instead to get you comfortable with the unit circle since that is a very important skill that will be needed in solving trig equations. cos(5π 6) cos. . (5 π 6) Solution. sin(−4π 3) sin. .

Solving Trigonometric Equations | Desmos. One way to solve a trigonometric equation is to graph the functions on each side of the equals sign,then find the point (s) of intersection. The example below shows you how to use this method. Example: Solve the equation sin (x)=1/2. To solve, start by turning on the two equations in boxes 3 & 4 below ...

Practice Questions on Trigonometry. Solve the following trigonometry problems. Prove that (sin α + cos α) (tan α + cot α) = sec α + cosec α. If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B. If sin θ + cos θ = √3, prove that tan θ + cot θ = 1. Evaluate: 2 tan 2 45° + cos 2 30° - sin 2 60°.