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RS Aggarwal Solutions Class 10 Chapter 5 Trigonometric Ratios
Exercise 5
Question 1:
Solution:
Given function: sin θ = √3/2
Let us first draw a right ∆ABC, ∠B = 90 degrees and ∠A = 𝜃
(where k is a positive)
We know that sin 𝜃 = BC/AC = (Perpendicular)/Hypotenuse = √3/2
By Pythagoras Theorem:
AC2 = AB2 + BC2
Or AB2 = AC2 – BC2 = 4k2 – 3k2 = k2
AB = k
Find other T-rations using their definitions:
Cos = AB/AC = 1/2
Tan 𝜃 = BC/𝐴𝐵 = √3
𝑐𝑜𝑠𝑒𝑐 𝜃 = 1/sin𝜃 = 2/√3
sec 𝜃 = 1/cos 𝜃 = 2
cot 𝜃 = 1/tan 𝜃 = 1/√3
Question 2:
Solution:
Given function: cos θ = 7/25
Draw a right ∆ABC, ∠B = 90 degrees and ∠A = θ
(where k is a positive)
We know that cos θ = AB/AC = Base/Hypotenuse = 7/25
By Pythagoras Theorem:
AC2 = AB2 + BC2
Or BC2 = AC2 – AB2 = 625k2 – 49k2 = 576k2
AB = 24k
Find other T-rations using their definitions:
Sin 𝜃 = BC/AC = 24/25
Tan 𝜃 = BC/𝐴𝐵 = 24/7
𝑐𝑜𝑠𝑒𝑐 𝜃 = 1/sin𝜃 = 25/24
sec 𝜃 = 1/cos 𝜃 = 25/7
cot 𝜃 = 1/tan 𝜃 = 7/24
Question 3:
Solution:
Given function: tan θ=15/8
Draw a right ∆ABC, ∠B = 90 degrees and ∠A = θ
We know that tan θ = BC/AB = perpendicular/base = 15/8
(where k is a positive)
By Pythagoras Theorem:
AC2 = AB2 + BC2
= 64k2 + 225k2
= 289k2
AC = 17k
Find other T-rations using their definitions:
Sin 𝜃 = BC/AC = 15/17
cos 𝜃 = AB/AC = 8/17
𝑐𝑜𝑠𝑒𝑐 𝜃 = 1/sin𝜃 = 17/15
sec 𝜃 = 1/cos 𝜃 = 17/8
cot 𝜃 = 1/tan 𝜃 = 8/15
Question 4:
Solution:
Given function: cot θ= 2
Draw a right ∆ABC, ∠C = 90 degrees and ∠A = θ
We know that cot θ = AC/BC = base/perpendicular = 2/1
(where k is a positive)
By Pythagoras Theorem:
AB2 = BC2 + AC2
= k2 + 4k2
= 5k2
AB = k√5
Find other T-rations using their definitions:
Sin 𝜃 = BC/AB = k/(k√5) = 1/√5
cos 𝜃 = AC/AB = (2k)/(k√5) = 2/√5
tan θ = BC/AC = sinθ /cosθ = k/(2k) = 1/2
𝑐𝑜𝑠𝑒𝑐 𝜃 = 1/sin𝜃 = √5
sec 𝜃 = 1/cos 𝜃 = √5/2
Question 5:
Solution:
Given function: cosec θ = √10
Draw a right ∆ABC, ∠C = 90 degrees and ∠A = θ
We know that, cosecθ = AB/BC = hypotenuse/perpendicular = (k√10)/k
(where k is a positive)
By Pythagoras Theorem:
AB2 = AC2 + BC2
AB2 = AC2 – BC2
= 10k2 – k2
= 9k2
AC = 3k
Find other T-rations using their definitions:
Sin 𝜃 = BC/AB = 1/√10
cos 𝜃 = AC/AB = (3k)/(k√10) = 3/√10
tan θ = BC/AC = sinθ /cosθ = 1/3
secθ = AB/AC = 1/cosθ = √10/3
cotθ = AC/BC = 1/tanθ = 3
Question 6:
Solution:
Draw a triangle, ΔABC, Let ∠ACB = θ and ∠B = 90 degrees
Sin θ = (a2 – b2) / (a2 + b2)
AB = (a2 – b2)
AC = (a2 + b2)
By Pythagoras theorem:
BC = √[(a2 + b2)2 – (a2 – b2)2]
BC = √(4a2 b2)
or BC = 2ab
Find other T-rations using their definitions:
cos 𝜃 = base/hypotenuse = 2ab / (a2 + b2)
tan θ = perpendicular/base = (a2 – b2) / 2ab
cosec 𝜃 = 1/sin 𝜃 = (a2 + b2)/(a2 – b2)
sec θ = 1/cos 𝜃 = (a2 + b2)/2ab
cotθ = 1/tanθ = 2ab/(a2 – b2)
Question 7:
Solution:
Given: 15 cot A = 8
cotA = (8k)/(15k) = 1/tanA = AC/BC
Where k is any positive.
By Pythagoras theorem:
AB2 = BC2 + AC2
= (15k)2 + (8k)2
= 289k2
AB = 17k
Find other T-rations using their definitions:
sin A = perpendicular/ hypotenuse = (15k)/(17k) = 15/17
sec A = hypotenuse /base = 17/8
Question 8:
Solution:
Draw a triangle, ΔABC, Let ∠ACB = θ and ∠B = 90 degress
Sin A = perpendicular/ hypotenuse = 9/41
By Pythagoras theorem:
AC2 = AB2 + BC2
AB2 = AC2 – BC2
= 412 – 92
= 1600
AB = 40
Find other T-ratios using their definitions:
cos A = base/ hypotenuse = 40/41
tan A = perpendicular /base = 9/40
Question 9:
Solution:
cos θ = 0.6 = (6k)/(10k) = AC/AB
Where k is any positive.
By Pythagoras theorem:
AB2 = BC2 + AC2
BC2 = AB2 – AC2
= (10k)2 + (6k)2
= 64k2
BC = 8k
Find other T-rations using their definitions:
sin θ = perpendicular/ hypotenuse = 8/10
tan θ = perpendicular/base = 8/6
Now,
LHS = 5sinθ – 3tanθ
= 5(8/10) – 3(8/6)
= 4 – 3(4/3)
= 4(3) – 3(4)
= 12 – 12
= 0
=RHS
Hence proved.
Question 10:
Solution:
cosec θ = 2
or 1/sinθ = 2
(cosecθ is reciprocal of sin θ)
sin θ = 1/2
which implies θ = 30 degrees.
Find the values of cos θ and cot θ at θ = 30 degrees.
cos 30^0 = √3/2 and cot 30^0 = √3
Now,
LHS = cot 0 + sin0/(1 +cos0)
= √3 + 1/2 / (1 + √3/2)
=2
=RHS
Hence proved.
Question 11:
Solution:
Given: tan θ = 1/√7
tanθ = k/(k√7) = BC/AC
Where k is any positive.
By Pythagoras theorem:
AB2 = BC2 + AC2
= k2 + 7k2
AB = 2k√2
Find cosec θ and sec θ using their definitions:
cosec θ = AB/BC = 2k√2/k = 2√2
secθ = AB/AC = 2√2/√7
Now,
LHS =
= 48/64
=3/4
= RHS
Hence proved
Question 12.
Solution:
Given: tan θ = 20/21
tanθ = 20k/(21k)
Where k is any positive.
By Pythagoras theorem:
AC2 = AB2 + BC2
= 441k2 + 400k2
AC = 29k
Find sin θ and cos θ using their definitions:
sin θ = perpendicular/ hypotenuse = 20/29
cos 𝜃 = base/hypotenuse = 21/29
Now,
LHS =
=30/70
= 3/7
=RHS
Hence proved
Question 13:
Solution:
Given: secθ=5/4
sec θ = 5k/(4k) = 5/4 and cos θ = 4/5
Where k is any positive.
By Pythagoras theorem:
AC2 = AB2 + BC2
BC2 = AC2 – AB2
= 25k2 – 16k2
BC = 9k2 = 3k
Find sin θ, tan θ and cot θ using their definitions:
sin θ = perpendicular/ hypotenuse = 3/5
tan θ = perpendicular/base = 3/4
cotθ = 1/tanθ = 4/3
Now,
LHS =
=RHS
Question 14:
Solution:
Given: cotθ=3/4
or cotθ = 3k/4k
Where k is any positive.
By Pythagoras theorem:
AC2 = AB2 + BC2
= 9k2 +16k2
= 25k2
AC = 5k
Find sec θ and cosec θ using their definitions:
sec θ = hypotenuse/base = 5/3
cosec 𝜃 = hypotenuse/perpendicular = 5/4
Now,
Question 15:
Solution:
Given: sinθ=3/4
or sinθ=3k/4k
Where k is any positive.
By Pythagoras theorem:
AC2 = AB2 + BC2
16k2 = AB2 + 9k2
AB = √7
Find sec θ and cot θ using their definitions:
sec θ = hypotenuse/base = 4/√7
cotθ = base/perpendicular = √7/3
Now,
Question 16:
Solution:
Given: sinθ = a/b
or sinθ= ak/bk
Where k is any positive.
By Pythagoras theorem:
AC2 = AB2 + BC2
AB2 = AC2 – BC2
= b2 – a2b2−a2−−−−−−√
Find sec θ and tan θ using their definitions:
sec θ = hypotenuse/base = b/√(b2 – a2)
tan θ = perpendicular/base = a/√(b2 – a2)
Now,
LHS = sec θ + tan θ
= RHS
Question 17:
Solution:
Given: cosθ=3/5
cosθ = (3k)/(5k) = AC/AB
Where k is any positive.
By Pythagoras theorem:
AB2 = BC2 + AC2
BC = 4k
Find sin θ, tan θ and cot θ using their definitions:
sin θ = perpendicular/ hypotenuse = 4/5
tan θ = perpendicular/base = 4/3
cot θ = 1/tan θ = 3/4
Now,
LHS =
= 3/160
= RHS
Question 18:
Solution:
Given: tan θ = 4/3
tanθ = (4k)/(3k) = BC/AC
Where k is any positive.
By Pythagoras theorem:
AB2 = BC2 + AC2
AB2 = 16k2 + 9k2
AB = 5k
Find sin θ and cos θ using their definitions:
sin θ = perpendicular/ hypotenuse = 4/5
cos 𝜃 = base/hypotenuse = 3/5
Now,
LHS = sin θ + cos θ
= 4/5 + 3/5
= 7/5
= RHS
Question 19:
Solution:
Given: tan θ = a/b
tanθ = (ak)/(bk) = BC/AB
Where k is any positive.
By Pythagoras theorem:
AC2 = AB2 + BC2AC=a2+b2−−−−−−√
Find sin θ and cos θ using their definitions:
sin θ = perpendicular/ hypotenuse = a/(√(a2+b2)
cos 𝜃 = base/hypotenuse = b/(√(a2+b2)
Now,
LHS:
Question 20:
Solution:
Given: 3 tan θ = 4
or tan θ = 4/3
Question 21:
Solution:
Given: 3 cot θ = 2
or cot θ = 2/3
Where k is any positive.
By Pythagoras theorem:
AC2 = AB2 + BC2
AC2 = (2k)2 + (3k)2
AC2 = 4k2 + 9k2 = 13k2
AC = √13 k
Find sin θ and cos θ using their definitions:
sin θ = perpendicular/ hypotenuse = 3/√13
cos 𝜃 = base/hypotenuse =2/√13
Now,
Question 22:
Solution:
Given: 3cotθ=4
or cot θ = 4/3
Where k is any positive.
By Pythagoras theorem:
AC2 = AB2 + BC2
= 16 + 9
AC = 5 k
Find sin θ and cos θ using their definitions:
sin θ = perpendicular/ hypotenuse = 3/5
cos 𝜃 = base/hypotenuse = 4/5
Now,
Question 23:
Solution:
Given: sec θ = 17/8
or sec θ = 17k/8k
Where k is any positive.
By Pythagoras theorem:
AC2 = AB2 + BC2
BC2 = AC2 – AB2
= 289k2 – 64k2
= 225 k2
BC = 15 k
Question 24:
Solution:
Draw a triangle using given instructions:
From figure: Δ ABC and Δ ABD are right angled triangles
where AD = 10cm BC = CD = 4cm
BD = BC + CD = 8cm
By Pythagoras theorem:
AD2 = BD2 + AB2
(10)2 = (8)2 + AB2
100 = 64 + AB2
AB2 = 36 = (6)2
or AB = 6cm
Again,
AC2 = BC2 + AB2
AC2 = (4)2 + (6)2
AC2 = 16 + 36 = 52
or AC = √52 = 2√13 cm
(i) Find sin θ
sin θ = BC/AC = 4/2√13 = 2√13/13
(ii) Find cos θ
cosθ = AB/AC = 6/2√13 = 3/√13 = 3√13/13
Question 25:
Solution:
Draw a triangle using given instructions:
From figure: Δ ABC is a right angled triangle
By Pythagoras theorem:
AC2 = BC2 + AB2
AC = 25
(i) Find sin A
sin A = BC/AC = 7/25
(ii) Find cos A
cos A = AB/AC = 24/25
(iii) sin C = AB/AC = 24/25
(iv) cosC = BC/AC = 7/25
Question 26:
Solution:
Draw a triangle using given instructions:
From figure: Δ ABC is a right angled triangle
AB2 = AC2 + BC2
(29)2 = AC2 + (21)2
841 = AC2 + 441
AC2 = 400
or AC = 20
Find sin θ and cos θ:
sinθ = AC/AB = 20/29
cosθ = BC/AB = 21/29
Now:
LHS = cos2θ – sin2θ
= (21/29)2 – (20/29)2
= 41/841
= RHS
Hence proved.
Question 27:
Solution:
From figure: Δ ABC is a right angled triangle
AC2 = BC2 + AB2
AC2 = (5)2 + (12)2
AC2 = 25 + 144
AC2 = 169 = (13)2
or AC = 13
Now from figure,
i. cos A = AB/AC = 12/13
ii. cosec A = AC/BC = 13/5
iii. cos C = BC/AC = 5/13
iv. cosec C = AC/AB = 13/12
Hence proved.
Question 28:
Solution:
Given: sinα = k/(2k) = BC/AB
Where k is any positive.
By Pythagoras theorem:
AB2 = BC2 + AC2
AC2 = AB2 – BC2
= (2k)2 – (k)2
= 3k2
or AC= k√3
Find cos α:
cos α = base/hypotenuse = √3/2
Now,
Question 29:
Solution:
Given: tan A = BC/AB = k/(k√3)
Where k is any positive.
By Pythagoras theorem:
AC2 = BC2 + AB2
= (k)2 – (√3 k)2
= k2 + 3k2
= 4k2
or AC= 2k
Find sin A, cos A, sin C and cos C
sin A = BC/AC = k/(2k) = 1/2
cos A = AB/AC = (k√3)/(2k) = √3/2
sin C = AB/AC = (k√3)/(2k) = √3/2
cos C = BC/AC = k/(2k) = ½
(i) sinA cosC + cosA sinC = 1
LHS = sinA cosC + cosA sinC = (1/2)(1/2) + ( √3/2)( √3/2)
= 1/4 + 3/4
= 4/4
= 1
RHS = LHS
(ii) cosA cosC – sinA sinC = 0
LHS = cosA cosC – sinA sinC
= (1/2)(√3/2) – (1/2)(√3/2)
= (√3/4) – (√3/4)
= 0
=RHS
Question 30:
Solution:
Consider two right triangles XAY and WBZ such that sin A = sin B
To Prove: ∠A = ∠B
From figures:
sin A = XY/XA and sin B = WZ / WB
XY/XA = WZ / WB = k (say)
or XY/WZ = XA/ WB …(1)
sin A = sin B (Given)
We have,
XY = WZ k and XA = WB k …(2)
By Pythagoras: Apply on both the triangles
WB2 = WZ2 + BZ2
BZ2 = WB2 – WZ2
And,
XA2 = XY2 + AY2
AY2 = XA2 – XY2
Question 31:
Solution:
Consider ΔABC to be a right angled triangle.
angle C = 90 degree
tan A = BC/AC and
tan B = AC/BC
Given: tanA = tanB
So, BC/AC = AC/BC
BC2 = AC2
BC = AC
Which implies, ∠ A = ∠ B (using triangle opposite and equal angles property)
Question 32:
Solution:
Consider ΔABC to be a right angled triangle at B.
angle C = 90 degree
Given: tan A = 1 …(1)
tan A = 1 = BC/AB
AB = BC
Again, tan A = sin A/cos A
sin A = cos A …using (1)
By Pythagoras theorem:
AC2 = BC2 + AB2
AC2 = 2BC2
(AC/BC)2 = 2
Or AC/BC = √2
cosecA = √2
or sin A = 1/√2
and cosA = 1/√2
Now,
2 sinA cosA = 2(1/√2)( 1/√2)
= 2(1/2)
= 1
= RHS
Question 33:
Solution:
Δ PQR is a right angled triangle.
By Pythagoras theorem:
PR2 = RQ2 + PQ2
(x + 2)2 = x2 + PQ2
PQ2 = 4 + 4x
or PQ = 2√(x+1)
Now,
cot ϕ = QR/PQ = x/2(√(x+1)
tan θ = QR/PQ = x/2(√(x+1)
(i) √(x+1) cot ϕ = √(x+1) {x/2(√(x+1)} = x/2
(ii) √(x^3 + x^ 2) tan θ = √(x^3 + x^ 2) {x/2(√(x+1)} = x2/2
(iii) cos θ = PQ/PR = 2(√(x+1) / (x+2)
Question 34:
(Using trig property: sin2 + cos2 A = 1)
= 1 – 1
= 0
Question 35:
Complete RS Aggarwal Solutions Class 10
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