1. In ∆ABC right angled at B, AB = 24 cm, BC = 7 m. Determine
(i) sin A, cos A
(ii) sin C, cos C
Solution:
(i)
Applying Pythagoras theorem for ∆ABC,
we obtain AC2 = AB2 + BC2
= (24 cm)2 + (7 cm)2
= (576 + 49) cm2
= 625 cm2
∴ AC = √(625)cm = 25 cm
We know that,
sinA = Opposite side/hypotenuse
sinA = BC/AC = 7/25
cosA = Adjacent side/hypotenuse
cosA = AB/AC = 24/25
(ii)
sinC = Opposite side/hypotenuse
= 24/25
cosC = Adjacent side/hypotenuse
= 7/25
2. In the given figure find tan P − cot R
Solution:
Applying Pythagoras theorem for ∆PQR,
we obtain PR2 = PQ2 + QR2
(13 cm)2 = (12 cm)2 + QR2
169 cm2 = 144 cm2 + QR2
25 cm2 = QR2
QR = 5 cm
tanP = opposite side/Adjacent side
tanP = 5/12
cotR = adjacent side/Opposite side
cotR = 5/12
tan P − cot R = 5/12 – 5/12 = 0
3. If sin A = ¾ , calculate cos A and tan A.
Solution:
Let ∆ABC be a right-angled triangle, right-angled at point B.
Given that, Let BC be 3k.
sinA = 3/4
cosA = 3/4
Therefore, AC will be 4k, where k is a positive integer.
Applying Pythagoras theorem in ∆ABC,
we obtain AC2 = AB2 + BC2
(4k)2 = AB2 + (3k)2
16k2 − 9k2 = AB2
7k2 = AB2
AB = √7k
4. Given 15 cot A = 8. Find sin A and sec A
Solution:
Consider a right-angled triangle, right-angled at B.
cotR = adjacent side/Opposite side
cotR = AB/BC
It is given that,
cotA = 8/15
Let AB be 8k.
Therefore, BC will be 15k, where k is a positive integer.
Applying Pythagoras theorem in ∆ABC,
we obtain AC2 = AB2 + BC2
= (8k) 2 + (15k) 2
= 64k2 + 225k2
= 289k2
AC = 17k
SinA = (opposite side)/(Hypotenuse side) = BC/AC = 15k/17k = 15/17
secA = hypotenuse/(adjacent side) = AC/AB = 15k/8k = 15/8
5. Given sec θ = 13/12 , calculate all other trigonometric ratios
Solution:
Consider a right-angle triangle ∆ABC, right-angled at point B.
secθ = hypotenuse/(adjacent side) = AC/AB = 13/12
If AC is 13k, AB will be 12k, where k is a positive integer.
Applying Pythagoras theorem in ∆ABC, we obtain
(AC)2 = (AB)2 + (BC)2
(13k)2 = (12k)2 + (BC)2
169k2 = 144k2 + BC2
25k2 = BC2
BC = 5k
sinθ = opposie side to θ)/hypotenuse = BC/AC = 5k/13k = 5/13
cosθ = (adjacent side to θ)/hypotenuse = AB/AC = 12k/13k = 12/13
tanθ = opposite side tp θ/adjacent side to θ = BC/AB = 5k/12k = 5/12
cotθ = adjacent side to θ/opposite side to θ = AB/BC = 12k/5k = 12/5
cosecθ = hypotenuse/adjacent side to θ = AC/BC = 13k/12k = 13/12
6. If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Solution:
Let us consider a triangle ABC in which CD ⊥ AB.
It is given that ,
cos A = cos B
AD/AC = BD/BC ————(1)
We have to prove ∠A = ∠B. To prove this, let us extend AC to P such that BC = CP.
From equation (1), we obtain
AD/BD = AC/BC
⇒AD/BD = AC/CP [By construction we have BC = CP]————(2)
By using the converse of B.P.T, CD||BP
⇒ ∠ACD = ∠CPB (Corresponding angles)———–(3)
∠BCD = ∠CBP (Alternate interior angles)———-(4)
By construction, we have BC = CP.
∴ ∠CBP = ∠CPB (Angle opposite to equal sides of a triangle)——(5)
From equations (3), (4), and (5), we obtain
∠ACD = ∠BCD————- (6)
In ∆CAD and ∆CBD, ∠ACD = ∠BCD [Using equation (6)]
∠CDA = ∠CDB [Both 90°]
Therefore, the remaining angles should be equal.
∴ ∠CAD = ∠CBD
⇒∠A = ∠B
7. If cot θ = 7/8 , evaluate
(i) [(1+sinθ)(1-sinθ)]/[(1+cosθ)(1-cosθ)]
(ii) cot²θ
Solution:
Let us consider a right triangle ABC, right-angled at point B.
cotθ = adjacent side to θ/opposite side to θ = BC/AB = 7/8
If BC is 7k, then AB will be 8k, where k is a positive integer.
Applying Pythagoras theorem in ∆ABC, we obtain
AC2 = AB2 + BC2
= (8k)2 + (7k)2
= 64k2 + 49k2
= 113k2
AC = √113k
sinθ = opposie side to θ)/hypotenuse = AB/AC = 8k/√113k = 8/√113
cosθ = (adjacent side to θ)/hypotenuse = BC/AC = 7k/√113k = 7/√113
(i) [(1+sinθ)(1-sinθ)]/[(1+cosθ)(1-cosθ)] = (1-sin²θ)/(1-cos²θ)
= [1 – (8/√113)²]/[1 – (7/√113)²]
= [1 – (64/113)]/[1 – (49/113)]
= [(49/113)]/[(64/113)]
= 49/64
(ii)cot²θ = cotθ² = (7/8)² = 49/64
8. If 3 cot A = 4, Check whether [(1-tan²A)/(1+tan²A)] = cos²A – sin²A or not
Solution:
It is given that 3cot A = 4 Or, cot A = 4/3
Consider a right triangle ABC, right-angled at point B.
cotA = adjacent side to A/opposite side to A = AB/BC = 4/3
If AB is 4k, then BC will be 3k, where k is a positive integer.
In ∆ABC, (AC)2 = (AB)2 + (BC)2
= (4k)2 + (3k)2
= 16k2 + 9k2
= 25k2
AC = 5k
sinA = opposie side to A)/hypotenuse = BC/AC = 3k/5k = 3/5
cosA = (adjacent side to A)/hypotenuse = AB/AC = 4k/5k = 4/5
tanθ = opposite side tp θ/adjacent side to θ = BC/AB = 3k/4k = 3/4
To check [(1-tan²A)/(1+tan²A)] = cos²A – sin²A
LHS, [(1-tan²A)/(1+tan²A)] = [1-(3/4)²/(1+3/4)²]
= [1-(9/16)]/[1+(9/16)]
= [(7/16)]/[(25/16)]
= 7/25,LHS
RHS, cos²A – sin²A
= (3/5)² – (4/5)²
= (9/25) – (16/25)
= [9-16]/25
= 7/25,RHS
Thus, LHS = RHS
Hence, [(1-tan²A)/(1+tan²A)] = cos²A – sin²A
9. In ∆ABC, right angled at B. If tanA = 1/√3 , find the value of
(i) sin A cos C + cos A sin C
(ii) cos A cos C − sin A sin C
Solution:
Given tanA = 1/√3
i.e., BC/AB = 1/√3
If BC is k, then AB will be , where k is a positive integer.
In ∆ABC, AC2 = AB2 + BC2
AC2 = (√3k)2 + k2
AC2 = 3k2 + k2 = 4k2
AC = √(4k2) = 2k
sinA= opposie side to A)/hypotenuse = BC/AC = k/2k = 1/2
cosA = (adjacent side to A)/hypotenuse = AB/AC = √3k/2k = √3/2
sinC= opposie side to C/hypotenuse = AB/AC = √3k/2k = √3/2
cosC = (adjacent side to C)/hypotenuse = BC/AC = k/2k = 1/2
To find, (i) sin A cos C + cos A sin C
= 1/2 x 1/2 + √3/2 x √3/2
= 1/4 + 3/4
= 4/4
= 1
to find (ii) cos A cos C − sin A sin C
= √3/2 x 1/2 – 1/2 x √3/2
= √3/4 – √3/4
= 0
10. In ∆PQR, right angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
Solution:
Given that, PR + QR = 25
PQ = 5
Let PR be x.
Therefore, QR = 25 − x
Applying Pythagoras theorem in ∆PQR, we obtain
PR2 = PQ2 + QR2
x2 = (5)2 + (25 − x)2
x2 = 25 + 625 + x2 − 50x
50x = 650
x = 13 Therefore, PR = 13 cm , QR = (25 − 13) cm = 12 cm
sinP = opposie side to P)/hypotenuse = RQ/PR = 12/13
cosP = (adjacent side to P)/hypotenuse = PQ/PR = 5/13
tanP = opposite side tp P/adjacent side to P = RQ/PQ = 12/5
11. State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
(ii) sec A = 12/5 for some value of angle A.
(iii) cos A is the abbreviation used for the cosec of angle A.
(iv) cot A is the product of cot and A (v) sin θ = 4/3 , for some angle θ
Solution:
(i) Consider a ∆ABC, right-angled at B.
tanA = opposite side tp A/adjacent side to A = 12/5
But 12/5 > 1
Therefore, tanA > 1
So, tan A < 1 is not always true. Hence, the given statement is false
(ii) secA = 12/5
sec A = AC/AB = 12/5
Let AC be 12k, AB will be 5k, where k is a positive integer.
Applying Pythagoras theorem in ∆ABC, we obtain
AC2 = AB2 + BC2
(12k)2 = (5k)2 + BC2
144k2 = 25k2 + BC2
BC2 = 119k2
BC = 10.9k
It can be observed that for given two sides AC = 12k and AB = 5k,
BC should be such that,
AC − AB < BC < AC + AB
12k − 5k < BC < 12k + 5k
7k < BC < 17 k
However, BC = 10.9k.
Clearly, such a triangle is possible and hence, such value of sec A is possible. Therefore, the given statement is true.
(iii) Abbreviation used for cosecant of angle A is cosec A. And cos A is the abbreviation used for cosine of angle A. Hence, the given statement is false.
(iv) cot A is not the product of cot and A. It is the cotangent of ∠A. Hence, the given statement is false.
(v) sin θ = 4/3
We know that in a right-angled triangle,
sinθ = opposie side to θ)/hypotenuse
In a right-angled triangle, hypotenuse is always greater than the remaining two sides. Therefore, such value of sin θ is not possible. Hence, the given statement is false.
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