TOPICS
Exercise - 8.1

Question-1 :-  In Δ ABC, right-angled at B, AB = 24 cm, BC = 7 cm.
Determine : (i) sin A, cos A  (ii) sin C, cos C

Solution :-
  Given :
  In Δ ABC, right-angled at B, AB = 24 cm, BC = 7 cm, AC = ?
  By Pythagoras Theorem :
  AC² = AB² + BC²
     = (24)² + (7)²
     = 576 + 49
     = 625
  AC = 25 cm
    NCERT Math
(i) sin A, cos A
  sin A = BC/AC = 7/25 
  cos A = AB/AC = 24/25

(ii) sin C, cos C
  sin C = AB/AC = 24/25
  cos C = BC/AC = 7/25
    

Question-2 :-  In Figure, find tan P – cot R. NCERT Math

Solution :-
  Given :
  In Δ PQR, right-angled at Q, PQ = 12 cm, PR = 13 cm, QR = ?
  By Pythagoras Theorem :
  QR² = PR² - PQ²
     = (13)² - (12)²
     = 169 - 144
     = 25
  QR = 5 cm
  
  tan P = QR/PQ = 5/12
  cot R = QR/PQ = 5/12
  tan P - cot R = 5/12 - 5/12 = 0
    

Question-3 :-  If sin A = 3/4, calculate cos A and tan A.

Solution :-
  Given :
  In Δ ABC, right-angled at B, 
  sin A = 3/4 = BC/AC
  AC = 4 cm, BC = 3 cm, AB = ?
  By Pythagoras Theorem :
  AB² = AC² - BC²
     = (4)² - (3)²
     = 16 - 9
     = 7
  AB = √7 cm 
  NCERT Math
  cos A = AB/AC = √7/4
  tan A = BC/AB = 3/√7
    

Question-4 :-  Given 15 cot A = 8, find sin A and sec A.

Solution :-
  Given :
  In Δ ABC, right-angled at B, 
  15 cot A = 8
     cot A = 8/15 = AB/BC
  AB = 8 cm, BC = 15 cm, AC = ?
  By Pythagoras Theorem :
  AC² = AB² + BC²
     = (8)² + (15)²
     = 64 + 225
     = 289
  AC = 17 cm 
  Ncert Math
  sin A = BC/AC = 15/17
  sec A = AC/AB = 17/8
    

Question-5 :-  Given sec θ = 13/12 calculate all other trigonometric ratios.

Solution :-
  Given :
  In Δ ABC, right-angled at B, 
  sec θ = 13/12 = AC/AB
  AB = 12 cm, BC = ?, AC = 13 cm
  By Pythagoras Theorem :
  BC² = AC² - AB²
     = (13)² - (12)²
     = 169 - 144
     = 25
  BC = 5 cm 
  ncert math
  sin A = BC/AC = 5/13
  cos A = AB/AC = 12/13
  tan A = BC/AB = 5/12
  cot A = AB/BC = 12/5
  cosec A = AC/BC = 13/5
    

Question-6 :-  If ∠ A and ∠ B are acute angles such that cos A = cos B, then show that ∠ A = ∠ B.

Solution :-
  Let us consider CD Perpendicular to AB
  ncert math
  Given :
  cos A = cos B
  AD/AC = BD/BC
  Let AD/BD = AC/BC = k
  AD = k BD .....(i)
  AC = k BC .....(ii)
  By using Pythagoras Theorem for Δ CAD and Δ CBD, we obtain :
  CD² = AC² - AD² .....(iii)
  CD² = BC² - BD² .....(iv)
  from equation (iii) & (iv), we obtain
  AC² - AD² = BC² - BD²
  (k BC)² - (k BD)² = BC² - BD²
  k² BC² - k² BD² = BC² - BD²
  k²(BC² - BD²) = BC² - BD²
  k² = 1
  k = ±1
  K = 1
  Putting this value in equation (ii), we obtain
  AC = BC
  So, ∠ A = ∠ B (Angles opposite to equal sides of a triangle)
    

Question-7 :-  If cot θ = 7/8, evaluate :
(i)ncert math
(ii) cot² θ

Solution :-
  Given :
  In Δ ABC, right-angled at B, 
  cot θ = 7/8 = AB/BC
  AB = 7 cm, BC = 8 cm, AC = ?
  By Pythagoras Theorem :
  AC² = AB² + BC²
     = (7)² + (8)²
     = 49 + 64
     = 113
  AC = √113 cm 
  ncert math
  sin θ = BC/AC = 8/√113
  cos θ = AB/Ac = 7/√113
(i)ncert math
(ii) cot² θ = (cos θ/sin θ)²
            = (7/√113 x √113/8)² 
            = (7/8)² 
            = 49/64
    

Question-8 :-  If 3 cot A = 4, check whether ncert math or Not.

Solution :-
  Given :
  In Δ ABC, right-angled at B, 
  3 cot A = 4
    cot A = 4/3 = AB/BC
  AB = 4 cm, BC = 3 cm, AC = ?
  By Pythagoras Theorem :
  AC² = AB² + BC²
     = (4)² + (3)²
     = 16 + 9
     = 25
  AC = 5 cm 
  ncert math
  sin A = BC/AC = 3/5
  cos A = AB/AC = 4/5
  tan A = BC/AB = 3/4
  
  Now, L.H.S
 ncert math
  R.H.S
  cos² A - sin² A = (4/5)² - (3/5)² = 16/25 - 9/25 = 7/25
  L.H.S = R.H.S
    

Question-9 :-  In triangle ABC, right-angled at B, if tan A = 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 :
  In Δ ABC, right-angled at B, 
  tan A = 1/√3 = BC/AB
  AB = √3 cm, BC = 1 cm, AC = ?
  By Pythagoras Theorem :
  AC² = AB² + BC²
     = (√3)² + (1)²
     = 3 + 1
     = 4
  AC = 2 cm 
  ncert math
  sin A = BC/AC = 1/2
  cos A = AB/AC = √3/2
  sin C = AB/AC = √3/2
  cos C = BC/AC = 1/2

(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

(ii) cos A cos C – sin A sin C
  = √3/2 x 1/2 - 1/2 x √3/2
  = √3/4 - √3/4
  = 0
    

Question-10 :-  In Δ 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 :
  In Δ PQR, right-angled at Q, 
  PR + QR = 25 cm and PQ = 5 cm
  PR = 25 - QR
  By Pythagoras Theorem :
  PR² = PQ² + QR²
  (25 - QR)² = (5)² + QR²
  25² + QR² - 50 QR = 25 + QR²
  625 - 50 QR = 25
  -50 QR = 25 - 625
  -50 QR = -600
  QR = 600/50
  QR = 12 cm
  Now, PR = 25 - QR = 25 - 12 = 13 cm
  ncert math
  sin P = QR/PR = 12/13
  cos P = PQ/PR = 5/13
  tan P = QR/PQ = 12/5
    

Question-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 cosecant 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 Δ ABC, right-angled at B.
   ncert math
  tan A = 10/5
  But 10/5 > 1
  tan A > 1
  So, tan A < 1 in not always true.
  Hence, the given statement is false.
    
(ii) sec A = 12/5 
  AC/AB = 12/5
  In Δ ABC, right-angled at B, 
  AC = 12 cm, AB = 5 cm, BC = ?
  By Pythagoras Theorem :
  BC² = AC² - AB²
     = (12)² - (5)²
     = 144 - 25
     = 119
  BC = 10.9 
  ncert math
  It can be observe that for give two sides AC = 12 and AB = 5
  BC should not be such that,
  (AC - AB) < BC < (AC + AB)
  (12 - 5) < BC < (12 + 5)
  7 < BC < 17
  However, BC = 10.9. Clearly, such a triangle is possible and hence, 
  such value of sec A is possible.
  Hence 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 a product of cot and A. It is the cotangent of angle A.
  Hence, the given statement is false.  
    
(v) sin θ = 4/3 
  We know that in a Right angled Triangle.
  sin θ = Perpendicular/Hypotaneouss
  In a right angled triangle, hypotaneous 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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