TRIGONOMETRICAL RATIOS - Questions & Answers
EXERCISE 21(A)1. Multiple Choice Type : Choose the correct answer from the options given below.
(a) If sin A = 5/13, the value of tan A is :
(i) 5/12 (ii) 12/13 (iii) 12/5 (iv) 13/12
Answer:
Given: sin A = 5/13.
We know that sin A = Perpendicular / Hypotenuse.
Let Perpendicular = 5 and Hypotenuse = 13.
Using Pythagoras Theorem in a right-angled triangle:
(Hypotenuse)² = (Perpendicular)² + (Base)²
13² = 5² + (Base)²
169 = 25 + (Base)²
(Base)² = 169 - 25
(Base)² = 144
Base = √144
Base = 12.
Now, we need to find tan A.
tan A = Perpendicular / Base.
tan A = 5 / 12.
The correct option is (i).
Given: sin A = 5/13.
We know that sin A = Perpendicular / Hypotenuse.
Let Perpendicular = 5 and Hypotenuse = 13.
Using Pythagoras Theorem in a right-angled triangle:
(Hypotenuse)² = (Perpendicular)² + (Base)²
13² = 5² + (Base)²
169 = 25 + (Base)²
(Base)² = 169 - 25
(Base)² = 144
Base = √144
Base = 12.
Now, we need to find tan A.
tan A = Perpendicular / Base.
tan A = 5 / 12.
The correct option is (i).
(b) If tan A = 3/5, the value of sin² A + cos² A is :
(i) 9/25 (ii) 1 (iii) 9/16 (iv) 16/9
Answer:
We know the fundamental trigonometric identity:
sin² A + cos² A = 1.
This identity holds true for any angle A.
Therefore, regardless of the value of tan A, the value of sin² A + cos² A will always be 1.
The correct option is (ii).
We know the fundamental trigonometric identity:
sin² A + cos² A = 1.
This identity holds true for any angle A.
Therefore, regardless of the value of tan A, the value of sin² A + cos² A will always be 1.
The correct option is (ii).
(c) If cot A = 5/12, the value of cot² A - cosec² A is :
(i) 1 (ii) 2 (iii) -2 (iv) -1
Answer:
We know the fundamental trigonometric identity:
cosec² A - cot² A = 1.
Multiplying both sides by -1, we get:
-(cosec² A - cot² A) = -1
-cosec² A + cot² A = -1
cot² A - cosec² A = -1.
The correct option is (iv).
We know the fundamental trigonometric identity:
cosec² A - cot² A = 1.
Multiplying both sides by -1, we get:
-(cosec² A - cot² A) = -1
-cosec² A + cot² A = -1
cot² A - cosec² A = -1.
The correct option is (iv).
(d) In the given figure (each observation is in cm) tan C is :
(i) 3/5 (ii) 4/3 (iii) 3/4 (iv) 4/5
Answer:
From the given figure, triangle ABD is right-angled at D.
Hypotenuse AB = 26 and Base BD = 10.
Using Pythagoras Theorem in triangle ABD:
(AB)² = (AD)² + (BD)²
26² = (AD)² + 10²
676 = (AD)² + 100
(AD)² = 676 - 100
(AD)² = 576
AD = √576 = 24.
Now, consider the right-angled triangle ADC.
Perpendicular AD = 24 and Base DC = 32.
tan C = Perpendicular / Base
tan C = AD / DC
tan C = 24 / 32
Simplifying the fraction by dividing numerator and denominator by 8:
tan C = 3 / 4.
The correct option is (iii).
From the given figure, triangle ABD is right-angled at D.
Hypotenuse AB = 26 and Base BD = 10.
Using Pythagoras Theorem in triangle ABD:
(AB)² = (AD)² + (BD)²
26² = (AD)² + 10²
676 = (AD)² + 100
(AD)² = 676 - 100
(AD)² = 576
AD = √576 = 24.
Now, consider the right-angled triangle ADC.
Perpendicular AD = 24 and Base DC = 32.
tan C = Perpendicular / Base
tan C = AD / DC
tan C = 24 / 32
Simplifying the fraction by dividing numerator and denominator by 8:
tan C = 3 / 4.
The correct option is (iii).
(e) If 5 cos A = 3, the value of sec² A - tan² A is :
(i) 1 (ii) -1 (iii) 3/4 (iv) 4/3
Answer:
We know the fundamental trigonometric identity:
sec² A - tan² A = 1.
This identity holds true for any acute angle A.
Therefore, the value is always 1.
The correct option is (i).
We know the fundamental trigonometric identity:
sec² A - tan² A = 1.
This identity holds true for any acute angle A.
Therefore, the value is always 1.
The correct option is (i).
2. From the following figure, find the values of :
(i) sin A (ii) cos A (iii) cot A (iv) sec C (v) cosec C (vi) tan C
Answer:
From the given right-angled triangle ABC, it is right-angled at B.
Base BC = 4 and Perpendicular AB = 3 (with reference to angle C).
Using Pythagoras Theorem to find Hypotenuse AC:
(AC)² = (AB)² + (BC)²
(AC)² = 3² + 4²
(AC)² = 9 + 16
(AC)² = 25
AC = √25 = 5.
(i) sin A = Perpendicular to angle A / Hypotenuse = BC / AC = 4 / 5.
(ii) cos A = Base for angle A / Hypotenuse = AB / AC = 3 / 5.
(iii) cot A = Base for angle A / Perpendicular to angle A = AB / BC = 3 / 4.
(iv) sec C = Hypotenuse / Base for angle C = AC / BC = 5 / 4.
(v) cosec C = Hypotenuse / Perpendicular to angle C = AC / AB = 5 / 3.
(vi) tan C = Perpendicular to angle C / Base for angle C = AB / BC = 3 / 4.
From the given right-angled triangle ABC, it is right-angled at B.
Base BC = 4 and Perpendicular AB = 3 (with reference to angle C).
Using Pythagoras Theorem to find Hypotenuse AC:
(AC)² = (AB)² + (BC)²
(AC)² = 3² + 4²
(AC)² = 9 + 16
(AC)² = 25
AC = √25 = 5.
(i) sin A = Perpendicular to angle A / Hypotenuse = BC / AC = 4 / 5.
(ii) cos A = Base for angle A / Hypotenuse = AB / AC = 3 / 5.
(iii) cot A = Base for angle A / Perpendicular to angle A = AB / BC = 3 / 4.
(iv) sec C = Hypotenuse / Base for angle C = AC / BC = 5 / 4.
(v) cosec C = Hypotenuse / Perpendicular to angle C = AC / AB = 5 / 3.
(vi) tan C = Perpendicular to angle C / Base for angle C = AB / BC = 3 / 4.
3. From the following figure, find the values of :
(i) cos B (ii) tan C (iii) sin² B + cos² B (iv) sin B.cos C + cos B.sin C
Answer:
From the given right-angled triangle ABC, it is right-angled at A.
Hypotenuse BC = 17 and Side AC = 8.
Using Pythagoras Theorem to find the third side AB:
(BC)² = (AB)² + (AC)²
17² = (AB)² + 8²
289 = (AB)² + 64
(AB)² = 289 - 64
(AB)² = 225
AB = √225 = 15.
(i) cos B = Base for angle B / Hypotenuse = AB / BC = 15 / 17.
(ii) tan C = Perpendicular to angle C / Base for angle C = AB / AC = 15 / 8.
(iii) Using the trigonometric identity, sin² B + cos² B is always equal to 1.
(iv) Let us first find the individual ratios:
sin B = Perpendicular to B / Hypotenuse = AC / BC = 8 / 17.
cos C = Base for C / Hypotenuse = AC / BC = 8 / 17.
cos B = Base for B / Hypotenuse = AB / BC = 15 / 17.
sin C = Perpendicular to C / Hypotenuse = AB / BC = 15 / 17.
Substitute these into the expression: sin B.cos C + cos B.sin C
= (8/17) * (8/17) + (15/17) * (15/17)
= 64/289 + 225/289
= (64 + 225) / 289
= 289 / 289 = 1.
From the given right-angled triangle ABC, it is right-angled at A.
Hypotenuse BC = 17 and Side AC = 8.
Using Pythagoras Theorem to find the third side AB:
(BC)² = (AB)² + (AC)²
17² = (AB)² + 8²
289 = (AB)² + 64
(AB)² = 289 - 64
(AB)² = 225
AB = √225 = 15.
(i) cos B = Base for angle B / Hypotenuse = AB / BC = 15 / 17.
(ii) tan C = Perpendicular to angle C / Base for angle C = AB / AC = 15 / 8.
(iii) Using the trigonometric identity, sin² B + cos² B is always equal to 1.
(iv) Let us first find the individual ratios:
sin B = Perpendicular to B / Hypotenuse = AC / BC = 8 / 17.
cos C = Base for C / Hypotenuse = AC / BC = 8 / 17.
cos B = Base for B / Hypotenuse = AB / BC = 15 / 17.
sin C = Perpendicular to C / Hypotenuse = AB / BC = 15 / 17.
Substitute these into the expression: sin B.cos C + cos B.sin C
= (8/17) * (8/17) + (15/17) * (15/17)
= 64/289 + 225/289
= (64 + 225) / 289
= 289 / 289 = 1.
4. From the following figure, find the values of :
(i) cos A (ii) cosec A (iii) tan² A - sec² A (iv) sin C (v) sec C (vi) cot² C - 1/sin² C
Answer:
From the given figure, BD is perpendicular to AC.
In right-angled triangle ABD, AD = 3 and BD = 4.
Using Pythagoras Theorem to find AB:
(AB)² = (AD)² + (BD)²
(AB)² = 3² + 4² = 9 + 16 = 25
AB = √25 = 5.
In right-angled triangle BDC, BD = 4 and DC = 12.
Using Pythagoras Theorem to find BC:
(BC)² = (BD)² + (DC)²
(BC)² = 4² + 12² = 16 + 144 = 160
BC = √160 = 4√10.
(i) cos A = Base / Hypotenuse = AD / AB = 3 / 5.
(ii) cosec A = Hypotenuse / Perpendicular = AB / BD = 5 / 4.
(iii) tan² A - sec² A = -(sec² A - tan² A) = -1 (Using identity).
(iv) sin C = Perpendicular / Hypotenuse = BD / BC = 4 / 4√10 = 1 / √10.
(v) sec C = Hypotenuse / Base = BC / DC = 4√10 / 12 = √10 / 3.
(vi) cot² C - 1/sin² C = cot² C - cosec² C = -(cosec² C - cot² C) = -1.
From the given figure, BD is perpendicular to AC.
In right-angled triangle ABD, AD = 3 and BD = 4.
Using Pythagoras Theorem to find AB:
(AB)² = (AD)² + (BD)²
(AB)² = 3² + 4² = 9 + 16 = 25
AB = √25 = 5.
In right-angled triangle BDC, BD = 4 and DC = 12.
Using Pythagoras Theorem to find BC:
(BC)² = (BD)² + (DC)²
(BC)² = 4² + 12² = 16 + 144 = 160
BC = √160 = 4√10.
(i) cos A = Base / Hypotenuse = AD / AB = 3 / 5.
(ii) cosec A = Hypotenuse / Perpendicular = AB / BD = 5 / 4.
(iii) tan² A - sec² A = -(sec² A - tan² A) = -1 (Using identity).
(iv) sin C = Perpendicular / Hypotenuse = BD / BC = 4 / 4√10 = 1 / √10.
(v) sec C = Hypotenuse / Base = BC / DC = 4√10 / 12 = √10 / 3.
(vi) cot² C - 1/sin² C = cot² C - cosec² C = -(cosec² C - cot² C) = -1.
5. From the following figure, find the values of :
(i) sin B (ii) tan C (iii) sec² B - tan² B (iv) sin² C + cos² C
Answer:
From the given figure, AD is perpendicular to BC.
In right-angled triangle ABD, Hypotenuse AB = 13 and Base BD = 5.
Using Pythagoras Theorem to find Perpendicular AD:
(AB)² = (AD)² + (BD)²
13² = (AD)² + 5²
169 = (AD)² + 25
(AD)² = 169 - 25 = 144
AD = √144 = 12.
In right-angled triangle ADC, Perpendicular AD = 12 and Base DC = 16.
(i) sin B = Perpendicular / Hypotenuse = AD / AB = 12 / 13.
(ii) tan C = Perpendicular / Base = AD / DC = 12 / 16 = 3 / 4.
(iii) sec² B - tan² B = 1 (Using fundamental identity).
(iv) sin² C + cos² C = 1 (Using fundamental identity).
From the given figure, AD is perpendicular to BC.
In right-angled triangle ABD, Hypotenuse AB = 13 and Base BD = 5.
Using Pythagoras Theorem to find Perpendicular AD:
(AB)² = (AD)² + (BD)²
13² = (AD)² + 5²
169 = (AD)² + 25
(AD)² = 169 - 25 = 144
AD = √144 = 12.
In right-angled triangle ADC, Perpendicular AD = 12 and Base DC = 16.
(i) sin B = Perpendicular / Hypotenuse = AD / AB = 12 / 13.
(ii) tan C = Perpendicular / Base = AD / DC = 12 / 16 = 3 / 4.
(iii) sec² B - tan² B = 1 (Using fundamental identity).
(iv) sin² C + cos² C = 1 (Using fundamental identity).
6. Given : sin A = 3/5, find :
(i) tan A (ii) cos A
Answer:
Given: sin A = 3 / 5.
We know, sin A = Perpendicular / Hypotenuse.
Let Perpendicular = 3 and Hypotenuse = 5.
Using Pythagoras Theorem to find Base:
(Hypotenuse)² = (Perpendicular)² + (Base)²
5² = 3² + (Base)²
25 = 9 + (Base)²
(Base)² = 25 - 9 = 16
Base = √16 = 4.
(i) tan A = Perpendicular / Base = 3 / 4.
(ii) cos A = Base / Hypotenuse = 4 / 5.
Given: sin A = 3 / 5.
We know, sin A = Perpendicular / Hypotenuse.
Let Perpendicular = 3 and Hypotenuse = 5.
Using Pythagoras Theorem to find Base:
(Hypotenuse)² = (Perpendicular)² + (Base)²
5² = 3² + (Base)²
25 = 9 + (Base)²
(Base)² = 25 - 9 = 16
Base = √16 = 4.
(i) tan A = Perpendicular / Base = 3 / 4.
(ii) cos A = Base / Hypotenuse = 4 / 5.
7. From the following figure, find the values of :
Answer:
The figure and sub-questions for question 7 are not completely printed/visible in the provided scanned chapter pages, hence it cannot be solved.
The figure and sub-questions for question 7 are not completely printed/visible in the provided scanned chapter pages, hence it cannot be solved.
8. Given : cos A = 5/13
evaluate : (i) (sin A - cot A) / (2 tan A) (ii) cot A + 1/cos A
Answer:
Given: cos A = 5 / 13.
We know, cos A = Base / Hypotenuse.
Let Base = 5 and Hypotenuse = 13.
Using Pythagoras Theorem to find Perpendicular:
(Hypotenuse)² = (Perpendicular)² + (Base)²
13² = (Perpendicular)² + 5²
169 = (Perpendicular)² + 25
(Perpendicular)² = 169 - 25 = 144
Perpendicular = √144 = 12.
Now find the required ratios:
sin A = Perpendicular / Hypotenuse = 12 / 13.
tan A = Perpendicular / Base = 12 / 5.
cot A = Base / Perpendicular = 5 / 12.
(i) Value of (sin A - cot A) / (2 tan A):
Substitute the values:
= (12/13 - 5/12) / (2 * 12/5)
Calculate the numerator: (144 - 65) / 156 = 79 / 156.
Calculate the denominator: 24 / 5.
Divide numerator by denominator: (79 / 156) * (5 / 24)
= 395 / 3744.
(ii) Value of cot A + 1/cos A:
Substitute the values:
= 5/12 + 1/(5/13)
= 5/12 + 13/5
Take LCM of 12 and 5 which is 60:
= (25 + 156) / 60
= 181 / 60.
Given: cos A = 5 / 13.
We know, cos A = Base / Hypotenuse.
Let Base = 5 and Hypotenuse = 13.
Using Pythagoras Theorem to find Perpendicular:
(Hypotenuse)² = (Perpendicular)² + (Base)²
13² = (Perpendicular)² + 5²
169 = (Perpendicular)² + 25
(Perpendicular)² = 169 - 25 = 144
Perpendicular = √144 = 12.
Now find the required ratios:
sin A = Perpendicular / Hypotenuse = 12 / 13.
tan A = Perpendicular / Base = 12 / 5.
cot A = Base / Perpendicular = 5 / 12.
(i) Value of (sin A - cot A) / (2 tan A):
Substitute the values:
= (12/13 - 5/12) / (2 * 12/5)
Calculate the numerator: (144 - 65) / 156 = 79 / 156.
Calculate the denominator: 24 / 5.
Divide numerator by denominator: (79 / 156) * (5 / 24)
= 395 / 3744.
(ii) Value of cot A + 1/cos A:
Substitute the values:
= 5/12 + 1/(5/13)
= 5/12 + 13/5
Take LCM of 12 and 5 which is 60:
= (25 + 156) / 60
= 181 / 60.
9. Given : sec A = 29/21, evaluate : sin A - 1/tan A
Answer:
Given: sec A = 29 / 21.
We know, sec A = Hypotenuse / Base.
Let Hypotenuse = 29 and Base = 21.
Using Pythagoras Theorem to find Perpendicular:
(Hypotenuse)² = (Perpendicular)² + (Base)²
29² = (Perpendicular)² + 21²
841 = (Perpendicular)² + 441
(Perpendicular)² = 841 - 441 = 400
Perpendicular = √400 = 20.
Now find the required ratios:
sin A = Perpendicular / Hypotenuse = 20 / 29.
tan A = Perpendicular / Base = 20 / 21.
Therefore, 1/tan A = cot A = 21 / 20.
Evaluate: sin A - 1/tan A
= 20/29 - 21/20
Take LCM of 29 and 20 which is 580:
= (400 - 609) / 580
= -209 / 580.
Given: sec A = 29 / 21.
We know, sec A = Hypotenuse / Base.
Let Hypotenuse = 29 and Base = 21.
Using Pythagoras Theorem to find Perpendicular:
(Hypotenuse)² = (Perpendicular)² + (Base)²
29² = (Perpendicular)² + 21²
841 = (Perpendicular)² + 441
(Perpendicular)² = 841 - 441 = 400
Perpendicular = √400 = 20.
Now find the required ratios:
sin A = Perpendicular / Hypotenuse = 20 / 29.
tan A = Perpendicular / Base = 20 / 21.
Therefore, 1/tan A = cot A = 21 / 20.
Evaluate: sin A - 1/tan A
= 20/29 - 21/20
Take LCM of 29 and 20 which is 580:
= (400 - 609) / 580
= -209 / 580.
10. Given : tan A = 4/3, find : cosec A / (cot A - sec A)
Answer:
Given: tan A = 4 / 3.
We know, tan A = Perpendicular / Base.
Let Perpendicular = 4 and Base = 3.
Using Pythagoras Theorem to find Hypotenuse:
(Hypotenuse)² = (Perpendicular)² + (Base)²
(Hypotenuse)² = 4² + 3²
(Hypotenuse)² = 16 + 9 = 25
Hypotenuse = √25 = 5.
Now find the required ratios:
cosec A = Hypotenuse / Perpendicular = 5 / 4.
cot A = Base / Perpendicular = 3 / 4.
sec A = Hypotenuse / Base = 5 / 3.
Calculate the denominator (cot A - sec A):
= 3/4 - 5/3
Take LCM of 4 and 3 which is 12:
= (9 - 20) / 12 = -11 / 12.
Now divide cosec A by the denominator:
= (5/4) / (-11/12)
= (5/4) * (-12/11)
= -15 / 11.
Given: tan A = 4 / 3.
We know, tan A = Perpendicular / Base.
Let Perpendicular = 4 and Base = 3.
Using Pythagoras Theorem to find Hypotenuse:
(Hypotenuse)² = (Perpendicular)² + (Base)²
(Hypotenuse)² = 4² + 3²
(Hypotenuse)² = 16 + 9 = 25
Hypotenuse = √25 = 5.
Now find the required ratios:
cosec A = Hypotenuse / Perpendicular = 5 / 4.
cot A = Base / Perpendicular = 3 / 4.
sec A = Hypotenuse / Base = 5 / 3.
Calculate the denominator (cot A - sec A):
= 3/4 - 5/3
Take LCM of 4 and 3 which is 12:
= (9 - 20) / 12 = -11 / 12.
Now divide cosec A by the denominator:
= (5/4) / (-11/12)
= (5/4) * (-12/11)
= -15 / 11.
11. Given : 4 cot A = 3, find :
(i) sin A (ii) sec A (iii) cosec² A - cot² A.
Answer:
Given: 4 cot A = 3.
Therefore, cot A = 3 / 4.
We know, cot A = Base / Perpendicular.
Let Base = 3 and Perpendicular = 4.
Using Pythagoras Theorem to find Hypotenuse:
(Hypotenuse)² = (Perpendicular)² + (Base)²
(Hypotenuse)² = 4² + 3² = 16 + 9 = 25
Hypotenuse = √25 = 5.
(i) sin A = Perpendicular / Hypotenuse = 4 / 5.
(ii) sec A = Hypotenuse / Base = 5 / 3.
(iii) cosec² A - cot² A = 1 (According to fundamental trigonometric identity).
Given: 4 cot A = 3.
Therefore, cot A = 3 / 4.
We know, cot A = Base / Perpendicular.
Let Base = 3 and Perpendicular = 4.
Using Pythagoras Theorem to find Hypotenuse:
(Hypotenuse)² = (Perpendicular)² + (Base)²
(Hypotenuse)² = 4² + 3² = 16 + 9 = 25
Hypotenuse = √25 = 5.
(i) sin A = Perpendicular / Hypotenuse = 4 / 5.
(ii) sec A = Hypotenuse / Base = 5 / 3.
(iii) cosec² A - cot² A = 1 (According to fundamental trigonometric identity).
12. Given : cos A = 0.6; find all other trigonometrical ratios for angle A.
Answer:
Given: cos A = 0.6 = 6 / 10 = 3 / 5.
We know, cos A = Base / Hypotenuse.
Let Base = 3 and Hypotenuse = 5.
Using Pythagoras Theorem to find Perpendicular:
(Hypotenuse)² = (Perpendicular)² + (Base)²
5² = (Perpendicular)² + 3²
25 = (Perpendicular)² + 9
(Perpendicular)² = 25 - 9 = 16
Perpendicular = √16 = 4.
Now find the remaining ratios:
sin A = Perpendicular / Hypotenuse = 4 / 5 = 0.8.
tan A = Perpendicular / Base = 4 / 3.
cot A = Base / Perpendicular = 3 / 4 = 0.75.
sec A = Hypotenuse / Base = 5 / 3.
cosec A = Hypotenuse / Perpendicular = 5 / 4 = 1.25.
Given: cos A = 0.6 = 6 / 10 = 3 / 5.
We know, cos A = Base / Hypotenuse.
Let Base = 3 and Hypotenuse = 5.
Using Pythagoras Theorem to find Perpendicular:
(Hypotenuse)² = (Perpendicular)² + (Base)²
5² = (Perpendicular)² + 3²
25 = (Perpendicular)² + 9
(Perpendicular)² = 25 - 9 = 16
Perpendicular = √16 = 4.
Now find the remaining ratios:
sin A = Perpendicular / Hypotenuse = 4 / 5 = 0.8.
tan A = Perpendicular / Base = 4 / 3.
cot A = Base / Perpendicular = 3 / 4 = 0.75.
sec A = Hypotenuse / Base = 5 / 3.
cosec A = Hypotenuse / Perpendicular = 5 / 4 = 1.25.
13. In a right-angled triangle, it is given that A is an acute angle and tan A = 5/12.
Find the values of :
(i) cos A (ii) sin A (iii) (cos A + sin A) / (cos A - sin A)
Answer:
Given: tan A = 5 / 12.
We know, tan A = Perpendicular / Base.
Let Perpendicular = 5 and Base = 12.
Using Pythagoras Theorem to find Hypotenuse:
(Hypotenuse)² = (Perpendicular)² + (Base)²
(Hypotenuse)² = 5² + 12²
(Hypotenuse)² = 25 + 144 = 169
Hypotenuse = √169 = 13.
(i) cos A = Base / Hypotenuse = 12 / 13.
(ii) sin A = Perpendicular / Hypotenuse = 5 / 13.
(iii) Substitute the values into the expression:
= (12/13 + 5/13) / (12/13 - 5/13)
= ((12 + 5)/13) / ((12 - 5)/13)
= (17/13) / (7/13)
= (17/13) * (13/7)
= 17 / 7.
Given: tan A = 5 / 12.
We know, tan A = Perpendicular / Base.
Let Perpendicular = 5 and Base = 12.
Using Pythagoras Theorem to find Hypotenuse:
(Hypotenuse)² = (Perpendicular)² + (Base)²
(Hypotenuse)² = 5² + 12²
(Hypotenuse)² = 25 + 144 = 169
Hypotenuse = √169 = 13.
(i) cos A = Base / Hypotenuse = 12 / 13.
(ii) sin A = Perpendicular / Hypotenuse = 5 / 13.
(iii) Substitute the values into the expression:
= (12/13 + 5/13) / (12/13 - 5/13)
= ((12 + 5)/13) / ((12 - 5)/13)
= (17/13) / (7/13)
= (17/13) * (13/7)
= 17 / 7.
14. Given : sin θ = p/q, find cos θ + sin θ in terms of p and q.
Answer:
Given: sin θ = p / q.
We know, sin θ = Perpendicular / Hypotenuse.
Let Perpendicular = p and Hypotenuse = q.
Using Pythagoras Theorem to find Base:
(Hypotenuse)² = (Perpendicular)² + (Base)²
q² = p² + (Base)²
(Base)² = q² - p²
Base = √(q² - p²).
Now find cos θ:
cos θ = Base / Hypotenuse = √(q² - p²) / q.
We need to find the value of: cos θ + sin θ.
Substitute the respective values:
= (√(q² - p²) / q) + (p / q)
Since the denominator is same, combine the numerators:
= (√(q² - p²) + p) / q.
Given: sin θ = p / q.
We know, sin θ = Perpendicular / Hypotenuse.
Let Perpendicular = p and Hypotenuse = q.
Using Pythagoras Theorem to find Base:
(Hypotenuse)² = (Perpendicular)² + (Base)²
q² = p² + (Base)²
(Base)² = q² - p²
Base = √(q² - p²).
Now find cos θ:
cos θ = Base / Hypotenuse = √(q² - p²) / q.
We need to find the value of: cos θ + sin θ.
Substitute the respective values:
= (√(q² - p²) / q) + (p / q)
Since the denominator is same, combine the numerators:
= (√(q² - p²) + p) / q.
15. If cos A = 1/2 and sin B = 1/√2, find the value of :
(tan A - tan B) / (1 + tan A tan B). Here angles A and B are from different right triangles. (HOTS)
Answer:
First, let us solve for angle A:
Given: cos A = 1 / 2.
Base = 1, Hypotenuse = 2.
Perpendicular for A = √(2² - 1²) = √(4 - 1) = √3.
tan A = Perpendicular / Base = √3 / 1 = √3.
Next, let us solve for angle B:
Given: sin B = 1 / √2.
Perpendicular = 1, Hypotenuse = √2.
Base for B = √((√2)² - 1²) = √(2 - 1) = √1 = 1.
tan B = Perpendicular / Base = 1 / 1 = 1.
Now, substitute the values of tan A and tan B into the expression:
Expression = (tan A - tan B) / (1 + tan A tan B)
= (√3 - 1) / (1 + (√3 * 1))
= (√3 - 1) / (√3 + 1)
To rationalize the denominator, multiply numerator and denominator by (√3 - 1):
= ((√3 - 1) * (√3 - 1)) / ((√3 + 1) * (√3 - 1))
= (√3 - 1)² / ( (√3)² - 1² )
= (3 + 1 - 2√3) / (3 - 1)
= (4 - 2√3) / 2
Divide numerator terms by 2:
= 2 - √3.
First, let us solve for angle A:
Given: cos A = 1 / 2.
Base = 1, Hypotenuse = 2.
Perpendicular for A = √(2² - 1²) = √(4 - 1) = √3.
tan A = Perpendicular / Base = √3 / 1 = √3.
Next, let us solve for angle B:
Given: sin B = 1 / √2.
Perpendicular = 1, Hypotenuse = √2.
Base for B = √((√2)² - 1²) = √(2 - 1) = √1 = 1.
tan B = Perpendicular / Base = 1 / 1 = 1.
Now, substitute the values of tan A and tan B into the expression:
Expression = (tan A - tan B) / (1 + tan A tan B)
= (√3 - 1) / (1 + (√3 * 1))
= (√3 - 1) / (√3 + 1)
To rationalize the denominator, multiply numerator and denominator by (√3 - 1):
= ((√3 - 1) * (√3 - 1)) / ((√3 + 1) * (√3 - 1))
= (√3 - 1)² / ( (√3)² - 1² )
= (3 + 1 - 2√3) / (3 - 1)
= (4 - 2√3) / 2
Divide numerator terms by 2:
= 2 - √3.
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Quick Review Flashcards - Click to flip and test your knowledge!
Question
What does the word 'Trigonometry' literally mean?
Answer
Measurement of triangles.
Question
In a right-angled triangle, what is the side opposite the acute angle of reference called?
Answer
The perpendicular.
Question
In a right-angled triangle, what is the side adjacent to the acute angle of reference called?
Answer
The base.
Question
In a right-angled triangle, what is the side opposite to the right angle called?
Answer
The hypotenuse.
Question
What is the common Greek letter \theta called in trigonometry?
Answer
Theta.
Question
What is the common Greek letter \phi called in trigonometry?
Answer
Phi.
Question
What is the common Greek letter \alpha called in trigonometry?
Answer
Alpha.
Question
What is the common Greek letter \beta called in trigonometry?
Answer
Beta.
Question
What is the common Greek letter \gamma called in trigonometry?
Answer
Gamma.
Question
How is a 'trigonometrical ratio' defined?
Answer
The ratio between the lengths of a pair of two sides of a right-angled triangle.
Question
Which ratio is defined as \frac{\text{perpendicular}}{\text{hypotenuse}}?
Answer
sine (\sin).
Question
Which ratio is defined as \frac{\text{base}}{\text{hypotenuse}}?
Answer
cosine (\cos).
Question
Which ratio is defined as \frac{\text{perpendicular}}{\text{base}}?
Answer
tangent (\tan).
Question
Which ratio is defined as \frac{\text{base}}{\text{perpendicular}}?
Answer
cotangent (\cot).
Question
Which ratio is defined as \frac{\text{hypotenuse}}{\text{base}}?
Answer
secant (\sec).
Question
Which ratio is defined as \frac{\text{hypotenuse}}{\text{perpendicular}}?
Answer
cosecant (\text{cosec}).
Question
What unit is used for a trigonometrical ratio?
Answer
It has no unit as it is a real number.
Question
What is the reciprocal of \sin A?
Answer
\text{cosec } A.
Question
What is the reciprocal of \cos A?
Answer
\sec A.
Question
What is the reciprocal of \tan A?
Answer
\cot A.
Question
\tan A is equal to the ratio of which two other trigonometrical ratios?
Answer
\frac{\sin A}{\cos A}.
Question
\cot A is equal to the ratio of which two other trigonometrical ratios?
Answer
\frac{\cos A}{\sin A}.
Question
The notation \sin^2 A is equivalent to what expression?
Answer
(\sin A)^2.
Question
What is the value of \sin^2 A + \cos^2 A for any angle A?
Answer
1.
Question
Complete the identity: \sec^2 A - \tan^2 A = \dots
Answer
1.
Question
Complete the identity: \text{cosec}^2 A - \dots = 1
Answer
\cot^2 A.
Question
What is the value of \sin 30^\circ?
Answer
\frac{1}{2}.
Question
What is the value of \cos 30^\circ?
Answer
\frac{\sqrt{3}}{2}.
Question
What is the value of \tan 30^\circ?
Answer
\frac{1}{\sqrt{3}}.
Question
What is the value of \sin 45^\circ?
Answer
\frac{1}{\sqrt{2}}.
Question
What is the value of \cos 45^\circ?
Answer
\frac{1}{\sqrt{2}}.
Question
What is the value of \tan 45^\circ?
Answer
1.
Question
What is the value of \sin 60^\circ?
Answer
\frac{\sqrt{3}}{2}.
Question
What is the value of \cos 60^\circ?
Answer
\frac{1}{2}.
Question
What is the value of \tan 60^\circ?
Answer
\sqrt{3}.
Question
What is the value of \sin 0^\circ?
Answer
0.
Question
What is the value of \cos 0^\circ?
Answer
1.
Question
What is the value of \tan 0^\circ?
Answer
0.
Question
What is the value of \sin 90^\circ?
Answer
1.
Question
What is the value of \cos 90^\circ?
Answer
0.
Question
What is the status of \tan 90^\circ?
Answer
It is not defined (\infty).
Question
As the angle increases from 0^\circ to 90^\circ, the value of \sin \theta increases from 0 to _____.
Answer
1.
Question
As the angle increases from 0^\circ to 90^\circ, the value of \cos \theta _____ from 1 to 0.
Answer
Decreases.
Question
As the angle increases from 0^\circ to 90^\circ, the value of \tan \theta increases from 0 to _____.
Answer
\infty (not defined).
Question
Under what condition are two acute angles said to be complementary?
Answer
If their sum is 90^\circ.
Question
What is the complement of an angle \theta^\circ?
Answer
(90 - \theta)^\circ.
Question
Complete the formula: \sin(90^\circ - \theta) = \dots
Answer
\cos \theta.
Question
Complete the formula: \cos(90^\circ - \theta) = \dots
Answer
\sin \theta.
Question
Complete the formula: \tan(90^\circ - \theta) = \dots
Answer
\cot \theta.
Question
Complete the formula: \cot(90^\circ - \theta) = \dots
Answer
\tan \theta.
Question
Complete the formula: \sec(90^\circ - \theta) = \dots
Answer
\text{cosec } \theta.
Question
Complete the formula: \text{cosec}(90^\circ - \theta) = \dots
Answer
\sec \theta.
Question
If x and y are acute angles and \sin x = \cos y, what is the relation between x and y?
Answer
x + y = 90^\circ.
Question
When solving \sin 2A = 1, what degree value replaces 1 to find 2A?
Answer
90^\circ.
Question
What is the value of A if 2 \cos 3A = 1 and A is acute?
Answer
20^\circ.
Question
In a triangle ABC right-angled at B, what is \sec A in terms of sides?
Answer
\frac{AC}{AB}.
Question
In a triangle ABC right-angled at B, what is \text{cosec } C in terms of sides?
Answer
\frac{AC}{AB}.
Question
For standard angles between 0^\circ and 90^\circ, which ratio is the reciprocal of 2 at 30^\circ?
Answer
\sin 30^\circ (reciprocal of \text{cosec } 30^\circ = 2).
Question
True or False: \sin(A + B) = \sin A + \sin B.
Answer
False.
Question
True or False: Each trigonometrical ratio of an angle depends only on the size of the angle and not on the size of the triangle.
Answer
True.