10.Trigonometry

 Navodaya

Formula Sheet

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Exercise 11.1

1.In right angle triangle ABC, 8 cm, 15 cm and 17 cm are the lengths of AB, BC and CA respectively. Then, find out sin A, cos A and tan A.

Solution :

 
Question 2:The sides of a right angle triangle PQR are PQ = 7 cm, QR = 25cm and ∠Q = 900 respectively. Then find, tan P – tan R.Solution :

Question 3:In a right angle triangle ABC with right angle at B, in which a = 24units,b = 25 units and ∠BAC = θ. Then, find cos θ and tan θ.

Solution :
Question 4:If cos A =  then find sin A and tan A.

Solution :

Question 5:
If 3 tan A = 4, then find sin A and cos A.
Solution :
Question 6:If ∠A and ∠ X are acute angles such that cos A = cos x then show that ∠A = ∠X.Solution :

Question 7:Solution :

Question 8:

In a right angle triangle ABC, right angle is at B, if tan A =  then find the value of

1. sin A cos C + cos A sin C
2. cos A cos C – sin A sin C

Solution

 

11.1 8th (i)

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Exercise 11.2

Question 1:

Solution :

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Question 2:
Solution :

Question 3:
Evaluate sin 600 cos 300 + sin 300 cos 600. What is the value of sin(600 + 300). What can you conclude ?
Solution :

Question 4:
Is it right to say cos(600 + 300) = cos 600 cos300 – sin 600 sin 300.
Solution :

Question 5:
In right angle triangle ∆PQR, right angle is at Q and PQ = 6 cms  ∠RPQ = 600 Determine the lengths of QR and PR.
Solution :

Question 6:
In ∆XYZ, right angle is at Y, Y Z = x, and XZ = 2x then determine ∠YXZ and ∠YZX.
Solution :

Question 7:

Is it right to say that sin (A+ B) = sin A + sin B? Justify your answer.

Solution :

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Exercise 11.3:

Question 1:

Solution :

Question 2:

Solution :

Question 3:
If tan2A = cot(A – 180), where 2A is an acute angle. Find the value of A.
Solution :
Given
tan2A=cot(A -18°)
cot(90-2A)=cot(A -18°)
Since 90-2A and A-18° are both acute angles,
Therefore
90-2A= A-18°
3A=108
A=36°
Question 4:

If tan A = cot B where A and B are acute angles, prove that A+ B = 900
Solution :Given tan A=cot B
tan (90-A)=cot B
since (90-A) and B are acute angles so,
90° – A=B
A + B=90°
Thus proved.

Question 5:
If A, B and C are interior angles of a triangle ABC, then show that tan  = cot 
Solution :

Question 6:
Express sin 750 + cos 650 in terms of trigonometric ratios of angles between 00 and 450
Solution :

We can write sin 75° =cos (90°-75°)= cos 15°
Cos 65˚=sin(90˚-65˚)=sin 25°
Then, sin 75°+cos 65°= cos 15°+ sin 25°
15° and 25° are between 0° and 45°

Exercise 11.4:

Question 1:

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Question 2:

Solution :Question 3:

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Question 4:

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Question 5:

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Question 6:
Simplify secA (1 – sinA) (secA + tanA)
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Question 7:
Prove that (sinA + cosec)2 + (cosA+ secA)2 = 7 + tan2A+ cot2A
Solution :

Question 8:
Simplify (1 – cos θ) (1 + cosθ) (1 + cot2θ)
Solution :

Question 9:
If secθ + tan θ = p, then what is the value of secθ – tanθ?
Solution :

Question 10:

Solution 

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