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.