practical trigonometry problem solving desmos answers

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• Trigonometry Problems - sin, cos, tan, cot

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 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

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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