Rs Aggarwal 2018 Solutions for Class 10 Math Chapter 5 Trigonometric Ratios are provided here with simple step-by-step explanations. These solutions for Trigonometric Ratios are extremely popular among Class 10 students for Math Trigonometric Ratios Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2018 Book of Class 10 Math Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2018 Solutions. All Rs Aggarwal 2018 Solutions for class Class 10 Math are prepared by experts and are 100% accurate.
Page No 288:
Question 1:
If sin , find the value of all T-ratios of θ.
Answer:
Let us first draw a right ABC, right angled at B and .
Now, we know that sin = = = .
So, if AB = , then AC = 2k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC2 = AB2 + BC2
⇒ BC2 = AC2 AB2 = (2k)2 ()2
⇒ BC2 = 4k2 3k2 = k2
⇒ BC = k
Now, finding the other T-ratios using their definitions, we get:
cos = =
tan =
∴ cot = , cosec = and sec =
Page No 288:
Question 2:
If cos find the values of all T-ratios of θ.
Answer:
Let us first draw a right ABC, right angled at B and .
Now, we know that cos = = = .
So, if BC = 7k, then AC = 25k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC2 = AB2 + BC2
⇒ AB2 = AC2 BC2 = (25k)2 (7k)2.
⇒ AB2 = 625k2 49k2 = 576k2
⇒ AB = 24k
Now, finding the other trigonometric ratios using their definitions, we get:
sin = =
tan =
∴ cot = , cosec = and sec =
Page No 288:
Question 3:
If tan find the values of all T-ratios of θ.
Answer:
Let us first draw a right ABC, right angled at B and .
Now, we know that tan = = = .
So, if BC = 8k, then AB = 15k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC2 = AB2 + BC2 = (15k)2 + (8k)2
⇒ AC2 = 225k2 + 64k2 = 289k2
⇒ AC = 17k
Now, finding the other T-ratios using their definitions, we get:
sin = =
cos =
∴ cot = , cosec = and sec =
Page No 288:
Question 4:
If cot θ = 2, find the value of all T-ratios of θ.
Answer:
Let us first draw a right ABC, right angled at B and .
Now, we know that cot = = = 2.
So, if BC = 2k, then AB = k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC2 = AB2 + BC2 = (2k)2 + (k)2
⇒ AC2 = 4k2 + k2 = 5k2
⇒ AC = k
Now, finding the other T-ratios using their definitions, we get:
sin = =
cos =
∴ tan = , cosec = and sec =
Page No 288:
Question 5:
If cosec θ = , the find the values of all T-ratios of θ.
Answer:
Let us first draw a right ABC, right angled at B and .
Now, we know that cosec = = = .
So, if AC = ()k, then AB = k, where k is a positive number.
Now, by using Pythagoras theorem, we have:
AC2 = AB2 + BC2
⇒ BC2 = AC2 AB2 = 10k2 k2
⇒ BC2 = 9k2
⇒ BC = 3k
Now, finding the other T-ratios using their definitions, we get:
tan = =
cos =
∴ , cot = and sec =
Page No 288:
Question 6:
If , find the values of all T-ratios of .
Answer:
We have ,
As,
Also,
Now,
Also,
And,
Page No 288:
Question 7:
If , find the values of sinA and secA.
Answer:
We have,
As,
Also,
Page No 288:
Question 8:
If , find the values of cosA and tanA.
Answer:
We have ,
As,
Also,
Page No 289:
Question 9:
If cos θ = 0.6, show that (5 sin θ − 3 tan θ) = 0.
Answer:
Let us consider a right ABC right angled at B.
Now, we know that cos = 0.6 = =
So, if BC = 3k, then AC = 5k, where k is a positive number.
Using Pythagoras theorem, we have:
AC2 = AB2 + BC2
⇒ AB2 = AC2 BC2
⇒ AB2 = (5k)2 (3k)2 = 25k2 9k2
⇒ AB2 = 16k2
⇒ AB = 4k
Finding out the other T-ratios using their definitions, we get:
sin =
tan =
Substituting the values in the given expression, we get:
5 sin 3 tan
i.e., LHS = RHS
Hence proved.
Page No 289:
Question 10:
If cosec θ = 2, show that
Answer:
Let us consider a right ABC, right angled at B and .
Now, it is given that cosec = 2.
Also, sin =
So, if AB = k, then AC = 2k, where k is a positive number.
Using Pythagoras theorem, we have:
⇒ AC2 = AB2 + BC2
⇒ BC2 = AC2 AB2
⇒ BC2 = (2k)2 (k)2
⇒ BC2 = 3k2
⇒ BC = k
Finding out the other T-ratios using their definitions, we get:
cos =
tan =
cot =
Substituting these values in the given expression, we get:
i.e., LHS = RHS
Hence proved.
Page No 289:
Question 11:
If tan θ = then prove that
Answer:
Let us consider a right ABC, right angled at B and .
Now it is given that tan = = .
So, if AB = k, then BC = k, where k is a positive number.
Using Pythagoras theorem, we have:
AC2 = AB2 + BC2
⇒ AC2 = (k)2 + (k)2
⇒ AC2 = k2 + 7k2
⇒ AC = 2k
Now, finding out the values of the other trigonometric ratios, we have:
sin =
cos =
∴ cosec = and sec =
Substituting the values of cosec and sec in the given expression, we get:
i.e., LHS = RHS
Hence proved.
Page No 289:
Question 12:
If tan θ = , show that
Answer:
Let us consider a right ABC right angled at B and .
Now, we know that tan = =
So, if AB = 20k, then BC = 21k, where k is a positive number.
Using Pythagoras theorem, we get:
AC2 = AB2 + BC2
⇒ AC2= (20k)2 + (21k)2
⇒ AC2 = 841k2
⇒ AC = 29k
Now, sin = and cos =
Substituting these values in the given expression, we get:
∴ LHS = RHS
Hence proved.
Page No 289:
Question 13:
If , show that .
Answer:
We have,
Also,
Now,
Page No 289:
Question 14:
If , show that .
Answer:
Page No 289:
Question 15:
If , show that .
Answer:
Page No 289:
Question 16:
If , show that .
Answer:
Page No 289:
Question 17:
If , show that .
Answer:
Page No 289:
Question 18:
If tan θ = , show that (sin θ + cos θ) = .
Answer:
Let us consider a right ABC, right angled at B and .
Now, we know that tan = = .
So, if BC = 3k, then AB = 4k, where k is a positive number.
Using Pythagoras theorem, we have:
AC2 = AB2 + BC2 = (4k)2 + (3k)2
⇒ AC2 = 16k2 + 9k2 = 25k2
⇒ AC = 5k
Finding out the values of sin and cos using their definitions, we have:
sin =
cos =
Substituting these values in the given expression, we get:
(sin + cos ) =
i.e., LHS = RHS
Hence proved.
Page No 289:
Question 19:
If tan θ = , show that
Answer:
It is given that tan .
LHS =
Dividing the numerator and denominator by cos , we get:
(∵ tan )
Now, substituting the value of tan in the above expression, we get:
i.e., LHS = RHS
Hence proved.
Page No 289:
Question 20:
If 3 tan θ = 4, show that
Answer:
Let us consider a right ABC right angled at B and .
We know that tan = =
So, if BC = 3k, then AB = 4k, where k is a positive number.
Using Pythagoras theorem, we have:
AC2 = AB2 + BC2
⇒ AC2 = 16k2 + 9k2
⇒ AC2 = 25k2
⇒ AC = 5k
Now, we have:
sin =
cos =
Substituting these values in the given expression, we get:
i.e., LHS = RHS
Hence proved.
Page No 289:
Question 21:
If 3 cot θ = 2, show that
Answer:
It is given that cot .
LHS =
Dividing the above expression by sin , we get:
[∵ cot ]
Now, substituting the values of cot in the above expression, we get:
i.e., LHS = RHS
Hence proved.
Page No 289:
Question 22:
If , show that .
Answer:
Since, LHS=RHS
Hence, verified.
Page No 289:
Question 23:
If sec θ = then prove that .
Answer:
It is given that sec = .
Let us consider a right ABC right angled at B and .
We know that cos =
So, if BC = 8k, then AC = 17k, where k is a positive number.
Using Pythagoras theorem, we have:
AC2 = AB2 + BC2
⇒ AB2 = AC2 BC2 = (17k)2 (8k)2
⇒ AB2 = 289k2 64k2 = 225k2
⇒ AB = 15k.
Now, tan = and sin =
The given expression is .
Substituting the values in the above expression, we get:
∴ LHS = RHS
Hence proved.
Page No 289:
Question 24:
In the adjoining figure,
Answer:
In ABD,
Using Pythagoras theorem, we get
Again,
In ABC,
Using Pythagoras therem, we get
Now,
Page No 289:
Question 25:
In a ∆ABC, it is given that ∠B = 90°, AB = 24 cm and BC = 7 cm.
Figure
Find the value of
(i) sin A
(ii) cos A
(iii) sin C
(iv) cos C
Answer:
Using Pythagoras theorem, we get:
AC2 = AB2 + BC2
⇒ AC2= (24)2 + (7)2
⇒ AC2 = 576 + 49 = 625
⇒ AC = 25 cm
Now, for T-Ratios of ∠A, base = AB and perpendicular = BC
(i) sin A =
(ii) cos A =
Similarly, for T-Ratios of ∠C, base = BC and perpendicular = AB
(iii) sin C =
(iv) cos C =
Page No 290:
Question 26:
Given a ∆ABC, in which ∠C = 90°, ∠ABC = θ°, BC = 21 units, AB = 29 units.
Show that (cos2θ − sin2θ) =
Answer:
Using Pythagoras theorem, we get:
AB2 = AC2 + BC2
⇒ AC2 = AB2 BC2
⇒ AC2 = (29)2 (21)2
⇒ AC2 = 841 441
⇒ AC2 = 400
⇒ AC = = 20 units
Now, sin and cos =
cos2 sin2 =
Hence Proved.
Page No 290:
Question 27:
In a ∆ABC, right-angled at B, it is given that AB = 12 cm and BC = 5 cm.
Find the value of
(i) cos A
(ii) cosec A
(iii) cos C
(iv) coses C.
Answer:
Using Pythagoras theorem, we get:
AC2 = AB2 + BC2
⇒ AC2 = 122 + 52 = 144 + 25
⇒ AC2 = 169
⇒ AC = 13 cm
Now, for T-Ratios of ∠A, base = AB and perpendicular = BC
(i) cos A =
(ii) cosec A =
Similarly, for T-Ratios of ∠C, base = BC and perpendicular = AB
(iii) cos C =
(iv) cosec C =
Page No 290:
Question 28:
If , prove that .
Answer:
Page No 290:
Question 29:
In a ABC, B = 90 and tanA = . Prove that
(i) sinAcosC + cosAsinC = 1 (ii) cosAcosC sinAsinC = 0
Answer:
Now,
Page No 290:
Question 30:
If A and B are acute angles such that sinA = sinB, then prove that A = B.
Answer:
In ABC, C = 90
sinA = and
sinB =
As, sinA = sinB
=
BC = AC
So, A = B (Angles opposite to equal sides are equal)
Page No 290:
Question 31:
If and are acute angles such that tanA = tanB, the prove that .
Answer:
In ,
Page No 290:
Question 32:
In a right , right-angled at , if , then verify that .
Answer:
Page No 290:
Question 33:
In the figure of . Find
Answer:
In ,
Using Pythagoras theorem, we get
Now,
Page No 290:
Question 34:
If and , then prove that .
Answer:
Page No 290:
Question 35:
If and , prove that .
Answer:
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