Search for notes by fellow students, in your own course and all over the country.
Browse our notes for titles which look like what you need, you can preview any of the notes via a sample of the contents. After you're happy these are the notes you're after simply pop them into your shopping cart.
Document Preview
Extracts from the notes are below, to see the PDF you'll receive please use the links above
Solution 7
…… given
In right ∆ABC, ∠B = 90° and ∠A = θ
Let BC = c and AC =
Then, AB2 = AC2 - BC2 = c2 + d2 - c2 = d2
Solution 8
In right ∆ABC, ∠B = 90° and ∠A = θ
√3 tan θ = 1 ⇒ tan θ =
Let BC = k and AB = √3k
Then, AC2 = AB2 + BC2 = 3k2 + k2 = 4k2
Solution 9
In right ∆ABC, ∠B = 90° and ∠A = θ
4 tan θ = 3 ⇒ tan θ
Let BC = 3k and AB = 4k
Then, AC2 = AB2 + BC2 = 16k2 + 9k2 = 25k2
Solution 11
In right ∆ABC, ∠B = 90° and ∠A = θ
AC2 = AB2 + BC2 = b2 + a2
Solution 12
Solution 13
In right ∆ABC, ∠B = 90° and ∠A = θ
Let BC = k and AB = 2k
Then, AC2 = AB2 + BC2 = (4k2 + k2)= 5k2
Solution 14
Solution 18
Solution 19
Let BC = 3k and AC = 4k
Then, AB2 = AC2 - BC2 = (16k2 - 9k2)= 7k2
Solution 20(i)
In right ∆ABC, ∠B = 90° and ∠A = θ
Let BC = 4k and AB = 3k
Then, AC2 = AB2 + BC2 = (9k2 + 16k2)= 25k2
Solution 20(ii)
In right ∆ABC, ∠B = 90° and ∠A = θ
Let BC = 4k and AB = 3k
Then, AC2 = AB2 + BC2 = (9k2 + 16k2)= 25k2
Solution 21
Solution 22
In right ∆ABC, ∠B = 90° and tan A = 1
Let BC = k and AB = k
Then, AC2 = AB2 + BC2 = (k2 + k2)= 2k2
Solution 23
In right ∆ABC, ∠B = 90°
Then, BC2 = AC2 - AB2 = (17)2 - (8)2= 289 - 64 = 225
⇒ BC = √225 = 15cm
Therefore Length = 15cm and Breadth = 8cm
(i) The area of rect
...
ABCD =2(l + b) = 2(15 + 8) = 46cm
Solution 27
Solution 28
Solution 1
Given:
Let us draw a
ABC in which
B = 90o and
BAC =
Solution 2
Let us draw a
ABC in which
B = 90o and
By Pythagoras theorem, we have
BAC =
Solution 3
Solution 4
Solution 5
Let us draw a
ABC in which
B = 90o and
BAC =
Solution 6
Solution 10
Solution 15
Given:
Let us draw a triangle ABC in which
B = 90o and
A=
By Pythagoras theorem, we have
Solution 16
Given:
Let us draw a triangle ABC in which
By Pythagoras theorem, we have
B = 90o and
A=
Solution 17
Given:
Let us draw a triangle ABC in which
By Pythagoras theorem, we have
B = 90o and
A=
Solution 24
Solution 25
Solution 26
Solution 29
Consider two right triangles XAY and WBZ such that sin A = sin B
Solution 30
Consider two right triangles XAY and WBZ such that tan A = tan B
Trigonometric Ratios Exercise MCQ
Solution 1
Let BC = 8k and AB = 15k
Then, AC2 = AB2 + BC2 = (225k2 + 64k2) = 289k2
⇒ AC2 =289k2
Correct Option: C
Solution 2
tan θ
Let BC = √3k and AB = k
Then, AC2 = AB2 + BC2 = (k2 + 3k2) = 4k2
⇒ AC2 =4k2
Correct Option: A
Solution 3
cosec θ = √10
Let BC = k and AC = √10k
Then, AB2 = AC2 - BC2 = (10k2 - k2) = 9k2
⇒ AB2 =9k2
⇒ AB = √9k2 = 3k
Correct Option: D
Solution 4
Correct Option: B
Solution 5
Let AB = 4k and AC = 5k
Then, BC2 = AC2 - AB2 = (25k2 - 16k2) = 9k2
⇒ BC = √9k2 = 3k
Correct Option: A
Solution 6
Let AB = 4k and AC = 5k
Then, BC2 = AC2 - AB2 = (25k2 - 16k2) = 9k2
⇒ BC = √9k2 = 3k
Correct Option: A
Solution 7
Correct Option: D
Solution 8
We have
(tan θ + cot θ) = 5
⇒ (tan θ + cot θ)2 = 52
⇒ tan2 θ + cot2 θ + 2 tan θ cot θ = 25
⇒ tan2 θ + cot2 θ + 2 tan θ ×
⇒ tan2 θ + cot2 θ + 2 = 25
⇒ tan2 θ + cot2 θ = 23
Correct Option: D
Solution 9
We have,
(cosθ + sec θ) =
⇒ (cosθ + sec θ)2 =
⇒ cos2 θ + sec2 θ + 2 cos θ sec θ =
⇒ tan2 θ + cot2 θ + 2 cos θ ×
⇒ tan2 θ + cot2 θ + 2 =
⇒ tan2 θ + cot2 θ =
Correct Option: A
Solution 10
Let BC = 3k and AB = 4k
Then, AC2 = AB2 + BC2 = (4k2 + 3k2) = 25k2
Correct Option: B
Solution 11
Correct Option: C
Solution 12
Let AB = 2k and AC = 3k
Then, BC2 = AC2 - AB2 = (9k2 - 4k2) = 5k2
Correct Option: A
Solution 13
We have, sec θ + tan θ + 1 = 0 ⇒ sec θ + tan θ = -1
We know that,
(sec2 θ - tan2 θ) = 1
⇒ (sec θ - tan θ)(sec θ + tan θ) = 1
⇒ (sec θ - tan θ) × -1 =1
⇒ (sec θ - tan θ) = -1
Correct Option: B
Solution 14
cos A + cos2 A = 1…