If tan (A + B) = √3 and tan (A – B) = 1/√3, 0° < A + B ≤ 90°; A > B, then find A and B
Trigonometry is basically the study of the relationship between the angles and the sides of a triangle. It is one of the widely used topics of mathematics that is used in daily life. It involves operations on a right-angled triangle i.e. a triangle having one of the angles equal to 90°. There are some terms that we should know before going further. These terms are,
- Hypotenuse – It is the side opposite to the right angle in a right-angled triangle. It is the longest side of a right-angled triangle. In figure1 side, AC is the hypotenuse.
- Perpendicular – The perpendicular of a triangle, corresponding to a particularly acute angle θ is the side opposite to the angle θ. In figure1 side, AB is the perpendicular corresponding to angle θ.
- Base – It is the side adjacent to a particularly acute angle θ. In figure 1 side BC is the base corresponding to angle θ.
Trigonometric Functions
As earlier said, trigonometry depicts the relationship between the angles and sides of a right-angled triangle. This relationship is represented by standard ratios and is given as follows:
- Sine (sin) The sine of an angle θ is the ratio of the length of the perpendicular, corresponding to the angle θ, to the length of the hypotenuse of the triangle.
sin θ = perpendicular/hypotenuse =p/h
- Cosine (cos) The cosine of an angle θ is the ratio of the length of the base, corresponding to the angle θ, to the length of the hypotenuse of the triangle.
cos θ = base/hypotenuse=b/h
- Tangent (tan) The tangent of an angle θ is the ratio of the length of the perpendicular, corresponding to the angle θ, to the length of the base for the particular angle of the triangle.
tan θ = perpendicular/base=p/b
- Cotangent (cot) It is the reciprocal of tangent.
cot θ = 1/tan θ = base/perpendicular=b/p
- Secant (sec) It is the reciprocal of cosine.
sec θ = 1/cos θ = hypotenuse/base=h/b
- Cosecant (cosec) It is the reciprocal of sine.
cosec θ =1/sin θ = hypotenuse/perpendicular=h/p
Some of the trigonometric ratios along with some of the standard angles are given in the table below,
0^{°} | 30° | 45° | 60° | 90° | |
sin | 0 | 1/2 | 1/√2 | √3/2 | 1 |
cos | 1 | √3/2 | 1/√2 | 1/2 | 0 |
tan | 0 | 1/√3 | 1 | √3 | ∞ |
cot | ∞ | √3 | 1 | 1/√3 | 0 |
sec | 1 | 2/√3 | √2 | 2 | ∞ |
cosec | ∞ | 2 | √2 | 2/√3 | 1 |
If tan (A + B) = √3 and tan (A – B) = 1/√3, 0° < A + B ≤ 90°; A > B, then find A and B
Solution:
Given,
tan(A + B) = √3
Since, A + B ≤ 90°,
Therefore,
A ≤ 90° and B ≤ 90°
According to the table, the angle at which tan acquires the value √3, is at 60°.
Therefore,
tan(A + B) = √3 = tan 60°
tan(A + B) = tan 60°
A + B = 60° ⇢ (i)
tan(A – B) = 1/√3
According to the table, the angle at which tan acquires the value 1/√3, is at 30°. Therefore,
tan(A – B) = 1/√3 = tan 30°
tan(A – B) = tan 30°
A – B = 30° ⇢ (ii)
Adding equation (i) and (ii),
2A = 90°
A = 45°
Putting the value of A in equation (i),
45° + B = 60°
B = 60°- 45°
B = 15°
Therefore, the value of A and B that satisfy the given equation is 45° and 15°, respectively.
Similar Problems
Question 1: If 2 sin^{2} θ – 1= 0, and 0°< θ< 90° then find the values of the following
a. cos θ + cos 2θ
b. sin θ × sin 2θ
Solution:
2 sin^{2} θ – 1 = 0
2sin^{2} θ = 1
sin^{2 }θ = 1/2
sin θ = 1/√2
The acute angle for which the value of sin is 1/√2 = 45°. Therefore,
sin θ = 1/√2 = sin 45°
θ = 45°
Therefore,
a. cos θ + cos 2θ
cos 45° + cos 2.45°
cos 45° + cos 90° (putting the values from the table)
1/√2 + 0 = 1/√2
b. sin θ × sin 2θ
sin 45° × sin 2.45°
sin 45° × sin 90°
1/√2 × 1
1/√2
Question 2: Find A and B if sin(A – B) = 1/2 = cos(A + B) and A, B are acute angles.
Solution:
sin(A – B) = 1/2
Since A, B are acute angles,
Therefore A – B should also be acute angle
The acute angle for which the value of sin is 1/2 = 30°. Therefore,
sin(A – B) = sin 30° = 1/2
A – B = 30° ⇢ (i)
The acute angle for which the value of cos is 1/2 = 60°, therefore,
cos(A + B) = cos 60° = 1/2
A + B = 60° ⇢ (ii)
Adding equations (i) & (ii),
2A = 90°
A = 45°
Putting the value of A in equation (i), we get
45° – B = 30°
B = 45° – 30°
B = 15°
Question 3: Find 2 tan^{2} θ + cos^{2 }θ – 1, if sin θ = cos θ, where 0°< θ< 90°.
Solution:
The acute angle for which the value of cos and sin are equal =45°, therefore,
cos θ = sin θ = cos 45°
θ = 45°
Therefore,
2 tan^{2} θ + cos^{2} θ -1
= 2 tan 45° × tan 45° + cos 45° × cos 45° -1
= 2 × 1 × 1 + 1/√2 × 1/√2 -1
= 2 + 1/2 -1
= 3/2
Question 4: Find θ for 0°<θ <90°, where (2sin θ – 1)(sin θ – 2)=0.
Solution:
(2sin θ – 1)(sin θ – 2) = 0
By looking at the equation, to satisfy the equation, either
(2sin θ – 1) = 0
or, (sin θ – 2) = 0
sin θ = 1/2
sin θ = 2 (since sin cannot exceed 1, therefore sin θ – 2 cannot be 0.)
sin θ = 1/2 = sin 30°
θ =30°
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