Access Answers for SAT Maths Chapter 8 – Introduction to
Trigonometry
Exercise 8.1
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 squares of the hypotenuse side is equal to the sum of the squares of
the other two sides.
By applying 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 sine (or) Sin function is the equal to the ratio of length of the opposite side to the
hypotenuse side. So it becomes
Sin (A) = Opposite side /Hypotenuse = BC/AC = 7/25
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 right angled triangle, to find the side QR, apply the Pythagorean theorem
According to Pythagorean theorem,
In a right- angled triangle, the squares 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 /Adjacent side = QR/PQ = 5/12
Since cot function is the reciprocal of the tan function, the ratio of cot function becomes,
Cot (R) = Adjacent side/Opposite side = 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.
, Solution:
Let us assume a right-angled triangle ABC, right angled at B
Given: Sin A = 3/4
We know that, Sin function is the equal to the ratio of 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 squares 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 numerator and denominator, 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 numerator and denominator, 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.
Solution:
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, cot function is the equal to the ratio of length of the adjacent side to the opposite
side.
Trigonometry
Exercise 8.1
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 squares of the hypotenuse side is equal to the sum of the squares of
the other two sides.
By applying 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 sine (or) Sin function is the equal to the ratio of length of the opposite side to the
hypotenuse side. So it becomes
Sin (A) = Opposite side /Hypotenuse = BC/AC = 7/25
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 right angled triangle, to find the side QR, apply the Pythagorean theorem
According to Pythagorean theorem,
In a right- angled triangle, the squares 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 /Adjacent side = QR/PQ = 5/12
Since cot function is the reciprocal of the tan function, the ratio of cot function becomes,
Cot (R) = Adjacent side/Opposite side = 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.
, Solution:
Let us assume a right-angled triangle ABC, right angled at B
Given: Sin A = 3/4
We know that, Sin function is the equal to the ratio of 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 squares 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 numerator and denominator, 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 numerator and denominator, 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.
Solution:
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, cot function is the equal to the ratio of length of the adjacent side to the opposite
side.
