EXERCISE 8.1
In ∆ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:
(i) sin A, cos A
(ii) sin C, cos C
Solution:
In a given triangle ABC, right angled at B = ∠B = 90°
Given: AB = 24 cm and BC = 7 cm
According to the Pythagoras Theorem,
In a right-angled triangle, the square of the hypotenuse side is equal to the sum of the squares of the other two sides.
By applying the Pythagoras theorem, we get
AC2=AB2+BC2
AC2 = (24)2+72
AC2 = (576+49)
AC2 = 625cm2
AC = √625 = 25
Therefore, AC = 25 cm
(i) To find sin (A), cos (A)
We know that the sine (or) sin function is equal to the ratio of the length of the opposite side to the hypotenuse side. So it becomes
Sin (A) = Opposite side/Hypotenuse = BC/AC = 7/25
The cosine or cos function is equal to the ratio of the length of the adjacent side to the hypotenuse side, and it becomes,
Cos (A) = Adjacent side/Hypotenuse = AB/AC = 24/25
(ii) To find sin (C), cos (C)
Sin (C) = AB/AC = 24/25
Cos (C) = BC/AC = 7/25
2. In Fig. 8.13, find tan P – cot R
Solution:
In the given triangle PQR, the given triangle is right-angled at Q, and the given measures are:
PR = 13cm
PQ = 12cm
Since the given triangle is a right-angled triangle, to find the side QR, apply the Pythagorean theorem
According to the Pythagorean theorem,
In a right-angled triangle, the square of the hypotenuse side is equal to the sum of the squares of the other two sides.
PR2 = QR2 + PQ2
Substitute the values of PR and PQ
132 = QR2+122
169 = QR2+144
Therefore, QR2 = 169−144
QR2 = 25
QR = √25 = 5
Therefore, the side QR = 5 cm
To find tan P – cot R:
According to the trigonometric ratio, the tangent function is equal to the ratio of the length of the opposite side to the adjacent sides; the value of tan (P) becomes
tan (P) = Opposite side/Hypotenuse = QR/PQ = 5/12
Since the cot function is the reciprocal of the tan function, the ratio of the cot function becomes,
Cot (R) = Adjacent side/Hypotenuse = QR/PQ = 5/12
Therefore,
tan (P) – cot (R) = 5/12 – 5/12 = 0
Therefore, tan(P) – cot(R) = 0
3. If sin A = 3/4, Calculate cos A and tan A.
Soln:
Let us assume a right-angled triangle ABC, right-angled at B
Given: Sin A = 3/4
We know that the sin function is equal to the ratio of the length of the opposite side to the hypotenuse side.
Therefore, Sin A = Opposite side/Hypotenuse = 3/4
Let BC be 3k, and AC will be 4k
where k is a positive real number.
According to the Pythagoras theorem, the square of the hypotenuse side is equal to the sum of the squares of the other two sides of a right-angle triangle, and we get,
AC2=AB2 + BC2
Substitute the value of AC and BC
(4k)2=AB2 + (3k)2
16k2−9k2 =AB2
AB2=7k2
Therefore, AB = √7k
Now, we have to find the value of cos A and tan A
We know that,
Cos (A) = Adjacent side/Hypotenuse
Substitute the value of AB and AC and cancel the constant k in both the numerator and denominator, and we get
AB/AC = √7k/4k = √7/4
Therefore, cos (A) = √7/4
tan(A) = Opposite side/Adjacent side
Substitute the Value of BC and AB and cancel the constant k in both the numerator and denominator, and we get
BC/AB = 3k/√7k = 3/√7
Therefore, tan A = 3/√7
4. Given 15 cot A = 8, find sin A and sec A.
Soln:
Let us assume a right-angled triangle ABC, right-angled at B
Given: 15 cot A = 8
So, Cot A = 8/15
We know that the cot function is equal to the ratio of the length of the adjacent side to the opposite side.
Therefore, cot A = Adjacent side/Opposite side = AB/BC = 8/15
Let AB be 8k and BC will be 15k
Where k is a positive real number.
According to the Pythagoras theorem, the square of the hypotenuse side is equal to the sum of the squares of the other two sides of a right-angle triangle, and we get,
AC2=AB2 + BC2
Substitute the value of AB and BC
AC2 = (8k)2 + (15k)2
AC2 = 64k2 + 225k2
AC2 = 289k2
Therefore, AC = 17k
Now, we have to find the value of sin A and sec A
We know that,
Sin (A) = Opposite side/Hypotenuse
Substitute the value of BC and AC and cancel the constant k in both the numerator and denominator, and we get
Sin A = BC/AC = 15k/17k = 15/17
Therefore, sin A = 15/17
Since the secant or sec function is the reciprocal of the cos function, which is equal to the ratio of the length of the hypotenuse side to the adjacent side,
Sec (A) = Hypotenuse/Adjacent side
Substitute the Value of BC and AB and cancel the constant k in both the numerator and denominator, and we get
AC/AB = 17k/8k = 17/8
Therefore, sec (A) = 17/8
5. Given sec θ = 13/12, calculate all other trigonometric ratios
Solution:
We know that the sec function is the reciprocal of the cos function, which is equal to the ratio of the length of the hypotenuse side to the adjacent side
Let us assume a right-angled triangle ABC, right-angled at B
sec θ =13/12 = Hypotenuse/Adjacent side = AC/AB
Let AC be 13k and AB will be 12k
Where k is a positive real number.
According to the Pythagoras theorem, the square of the hypotenuse side is equal to the sum of the squares of the other two sides of a right-angle triangle, and we get,
AC2=AB2 + BC2
Substitute the value of AB and AC
(13k)2= (12k)2 + BC2
169k2= 144k2 + BC2
169k2= 144k2 + BC2
BC2 = 169k2 – 144k2
BC2= 25k2
Therefore, BC = 5k
Now, substitute the corresponding values in all other trigonometric ratios
So,
Sin θ = Opposite Side/Hypotenuse = BC/AC = 5/13
Cos θ = Adjacent Side/Hypotenuse = AB/AC = 12/13
tan θ = Opposite Side/Adjacent Side = BC/AB = 5/12
Cosec θ = Hypotenuse/Opposite Side = AC/BC = 13/5
cot θ = Adjacent Side/Opposite Side = AB/BC = 12/5