Mathematics Part ii Solutions Solutions for Class 9 Maths Chapter 8 Trigonometry are provided here with simple step-by-step explanations. These solutions for Trigonometry are extremely popular among class 9 students for Maths Trigonometry Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part ii Solutions Book of class 9 Maths Chapter 8 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part ii Solutions Solutions. All Mathematics Part ii Solutions Solutions for class 9 Maths are prepared by experts and are 100% accurate.
Page No 104:
Question 1:
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In the given Fig, R is the right angle of PQR. Write the following ratios.
(i) sin P (ii) cos Q (iii) tan P (iv) tan Q
Answer:
![](/img/site_content/ask-answer/loader.gif)
(i) sinP =
(ii) cosQ =
(iii) tanP =
(iv) tanQ =
Page No 104:
Question 2:
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In the right angled XYZ, XYZ = 90° and a, b, c are the lengths of the sides as shown in the figure. Write the following ratios,
(i) sin X (ii) tan Z (iii) cos X (iv) tan X.
Answer:
![](/img/site_content/ask-answer/loader.gif)
(i) sinX =
(ii) tanZ =
(iii) cosX =
(iv) tanX =
Page No 104:
Question 3:
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Answer:
![](/img/site_content/ask-answer/loader.gif)
(i) sin50º =
(ii) cos50º =
(iii) tan40º =
(iv) cos40º =
Page No 104:
Question 4:
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In the given figure, PQR = 90° , PQS = 90° , PRQ = andQPS = Write the following trigonometric ratios.
(i) sin, cos , tan
(ii) sin, cos, tan
Answer:
![](/img/site_content/ask-answer/loader.gif)
(i)
sin =
cos =
tan =
(ii)
sinθ =
cosθ =
tanθ =
Page No 112:
Question 1:
sin | |
|
|
||||||
cos | |||||||||
tan | 1 | |
|
Answer:
(i)
Now,
(ii)
Now,
(iii)
Now,
and
(iv)
Now,
(v)
Now,
(vi)
Now,
and
(vii)
Now,
and
(viii)
Now,
(ix)
Now,
and
The complete table is given below:
sinθ | |||||||||
cosθ | |||||||||
tanθ | 1 |
Page No 112:
Question 2:
Find the values of -
(i) 5 sin 30â° + 3 tan 45° (ii) (iii) 2sin 30° + cos 0° + 3sin 90°
(iv) (v) (vi) cos 60°× cos 30° + sin 60°× sin 30°
Answer:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Page No 112:
Question 3:
If sin = then find cos
Answer:
Thus, the value of cosθ is .
Page No 112:
Question 4:
If cos = then find sin
Answer:
Thus, the value of sinθ is .
Page No 113:
Question 1:
(C) sin = tan (90) (D) tan = tan (90â)
(C) (D) 1
(iv)
(A) 2 (B) 1 (C) 0 (D) 1
Answer:
(i)
We know, .
Also,
and
Hence, the correct answer is option (A).
(ii)
We know, sin90º = 1.
Hence, the correct answer is option (D).
(iii)
Hence, the correct answer is option (C).
(iv)
Hence, the correct answer is option (D).
Page No 113:
Question 2:
In right angled TSU, TS = 5, S = 90°, SU = 12 then find sin T, cos T, tan T. Similarly find sin U, cos U, tan U.
Answer:
![](/img/site_content/ask-answer/loader.gif)
In right âTSU,
TU2 = SU2 + TS2 (Pythagoras theorem)
⇒ TU2 = 122 + 52 = 144 + 25 = 169
⇒ TU2 = 132
⇒ TU = 13
Now,
Also,
Page No 113:
Question 3:
In right angled YXZ, X = 90°, XZ = 8 cm, YZ = 17 cm, find sin Y, cos Y, tan Y, sin Z, cos Z, tan Z.
Answer:
![](/img/site_content/ask-answer/loader.gif)
In right âYXZ,
YZ2 = XZ2 + XY2 (Pythagoras theorem)
⇒ XY2 = YZ2 − XZ2
⇒ XY2 = 172 − 82 = 289 − 64 = 225
⇒ XY2 = 152
⇒ XY = 15 cm
Now,
Also,
Page No 113:
Question 4:
In right angled LMN , if N = , M = , , find sin and tan Similarly, find ( sin2 ) and ( cos2 ).
Answer:
![](/img/site_content/ask-answer/loader.gif)
In right âLMN, ∠N = θ.
Let MN = 24k and LN = 25k.
Using Pythagoras theorem, we have
LN2 = LM2 + MN2
⇒ (25k)2 = LM2 + (24k)2
⇒ LM2 = 625k2 − 576k2 = 49k2
⇒ LM2 = (7k)2
⇒ LM = 7k
Now,
Also,
Page No 113:
Question 5:
Fill in the blanks.
(i) sin200 = cos
(ii) tan300 tan = 1
(iii) cos400 = sin
Answer:
(i)
(ii)
Now,
(iii)
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