Class 10 NCERT Math Solution

Unit-8(Examples)

Example-1 :- Given tan A = 4/3, find the other trigonometric ratios of the angle A.

Solution :-

  Give that:

  tan A = 4/3
  cot A = 1/tan A = 3/4
  sec² A = 1 + tan² A
         = 1 + (4/3)²
         = 1 + 16/9
         = 25/9
  sec A = 5/3
  cos A = 1/sec A = 3/5
  sin² A = 1 - cos² A
         = 1 - (3/5)²
         = 1 - 9/25
         = 16/25
  sin A = 4/5
  cosec A = 1/sin A = 5/4

Example-2 :- If ∠ B and ∠ Q are acute angles such that sin B = sin Q, then prove that ∠ B = ∠ Q.

Solution :-

  Let us consider two right triangles ABC and PQR where sin B = sin Q see in figure
   
  We have sin B = AC/AB
          sin Q = PR/PQ
  Then, AC/AB = PR/PQ
  Therefore, AC/PR = AB/PQ = k  .......(1)
  Now using Pythagoras theorem,
  
  From (1) and (2), we have
  AC/PR = AB/PQ = BC/QR
  Δ ACB ~ Δ PRQ and therefore, ∠ B = ∠ Q.

Example-3 :- Consider Δ ACB, right-angled at C, in which AB = 29 units, BC = 21 units and ∠ ABC = θ (see Figure). Determine the values of (i) cos² θ + sin² θ, (ii) cos² θ – sin² θ NCERT Math

Solution :-

  In Δ ACB, we have AB = 29 units and BC = 21 units
  AC² = AB² - BC² 
      = (29)² - (21)²
      = (29 - 21)(29 + 21)
      = 8 x 50
      = 400
  AC = 20 units
  so, sin θ = AC/AB = 20/29, cos θ = 21/29

  Now, 
(i) cos² θ + sin² θ 
  = (21/29)² + (20/29)²
  = 441/841 + 400/841
  = 841/841
  = 1

(ii)cos² θ – sin² θ
  = (21/29)² - (20/29)²
  = 441/841 - 400/841
  = 41/841

Example-4 :- In a right triangle ABC, right-angled at B, if tan A = 1, then verify that 2 sin A cos A = 1.

Solution :-

  
  In Δ ABC, tan A = BC/AB = 1 (see in figure)
  i.e., BC = AB
  Let AB = BC = k, where k is a positive number
  AC² = AB² + BC²
      = k² + k²
      = 2k²
  AC = √2k
  Therefore, sin A = BC/AC = k/√2k = 1/√2
             cos A = AB/AC = 1/√2
  so, 2 sin A . cos A = 2 (1/√2) x (1/√2) = 1
  which is the required value.

Example-5 :- In Δ OPQ, right-angled at P, OP = 7 cm and OQ – PQ = 1 cm (see Figure). Determine the values of sin Q and cos Q. NCERT Math

Solution :-

  In Δ OPQ, we have
  OQ² = OP² + PQ²
  (1 + PQ)² = OP² + PQ² 
  i.e., 1 + PQ² + 2PQ = OP² + PQ²
  i.e., 1 + 2PQ = 7²
  i.e., PQ = 24 cm and OQ = 1 + PQ = 25 cm
  so, sin Q = 7/25 and cos Q = 24/25

Example-6 :- In Δ ABC, right-angled at B, AB = 5 cm and ∠ ACB = 30° (see Figure). Determine the lengths of the sides BC and AC. NCERT Math

Solution :-

  To find the length of the side BC, we will choose the trigonometric ratio involving BC and the given side AB. 
  Since BC is the side adjacent to angle C and AB is the side opposite to angle C, therefore
  AB/BC = tan C
   5/BC = tan 30°
   5/BC = 1/√3
     BC = 5√3 cm
  To find the length of the side AC, we consider
  sin 30° = AB/AC
  1/2 = 5/AC
  AC = 10cm

Example-7 :- In Δ PQR, right-angled at Q (see Figure), PQ = 3 cm and PR = 6 cm. Determine ∠ QPR and ∠ PRQ. NCERT Math

Solution :-

  Given PQ = 3 cm and PR = 6 cm
  Therefore, PQ/PR = sin R 
  sin R = 3/6 = 1/2
  so, ∠ PRQ = 30°
      ∠ QPR = 60°

Example-8 :- If sin (A – B) = 1/2, cos (A + B) = 1/2, 0° < A + B ≤ 90°, A > B, find A and B.

Solution :-

  Since, sin (A – B) = 1/2, 
         sin (A - B) = sin 30°
  Therefore, A – B = 30° .....(1)
  Also, since cos (A + B) = 1/2, 
              cos (A + B) = 60°
  Therefore, A + B = 60° ......(2)
  Solving (1) and (2), 
  A - B + A + B = 30° + 60°
  2A = 90°
   A = 90°/2 
   A = 45°
  Put in (1) equation
  A - B = 30°
  45° - B = 30°
  -B = 30° - 45°
  -B = -15°
   B = 15° 
  we get : A = 45° and B = 15°.

Example-9 :- Evaluate : tan 65°/cot 25°

Solution :-

  We know : cot A = tan (90° – A) 
  So, cot 25° = tan (90° – 25°) = tan 65°
  i.e., tan 65°/cot 25° = tan 65°/tan 65° = 1

Example-10 :- If sin 3A = cos (A – 26°), where 3A is an acute angle, find the value of A.

Solution :-

  We are given that sin 3A = cos (A – 26°)..... (1) 
  Since sin 3A = cos (90° – 3A), 
  we can write (1) as cos (90° – 3A) = cos (A – 26°) 
  Since 90° – 3A and A – 26° are both acute angles, 
  Therefore, 90° – 3A = A – 26° which gives A = 29°

Example-11 :- Express cot 85° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.

Solution :-

  cot 85° + cos 75° 
= cot (90° – 5°) + cos (90° – 15°) 
= tan 5° + sin 15°

Example-12 :- Express the ratios cos A, tan A and sec A in terms of sin A.

Solution :-

  Since cos² A + sin² A = 1, 
  Therefore, cos² A = 1 – sin² A, 
  i.e., cos A = √1 - sin² A
  tan A = sin A/cos A = sin A/√1 - sin² A
  sec A = 1/cos A = 1/√1 - sin² A

Example-13 :- Prove that : sec A (1 – sin A)(sec A + tan A) = 1.

Solution :-