Introduction to Trigonometry Ex 8.1 Solutions
Question - 1 : - In ∆ ABC,right-angled at B, AB = 24 cm, BC = 7 cm. Determine:
Answer - 1 : -
(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 ofthe 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 theequal to the ratio of length of the opposite side to the hypotenuse side. So itbecomes
Sin (A) = Opposite side /Hypotenuse = BC/AC =7/25
Cosine or Cos function is equal to the ratio ofthe 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
Question - 2 : - In Fig. 8.13, findtan P – cot R
Answer - 2 : - 
Question - 3 : - If sin A = 3/4, Calculate cos A and tan A.
Answer - 3 : - 
Question - 4 : - Given 15 cot A = 8, find sin A and sec A.
Answer - 4 : - 
Question - 5 : - Given sec θ = 13/12 Calculate all othertrigonometric ratios
Answer - 5 : - 
Question - 6 : - If ∠A and ∠B are acute angles such that cos A =cos B, then show that ∠ A = ∠ B.
Answer - 6 : - 
Question - 7 : - If cot θ = 7/8,evaluate :
Answer - 7 : -
(i) (1 + sin θ)(1 – sin θ)/(1+cos θ)(1-cos θ)
(ii) cot_2 θ
Solution:
Let us assume a △ABC in which ∠B = 90° and ∠C = θ
Given:
cot θ = BC/AB = 7/8
Let BC = 7k and AB = 8k, where k is a positivereal number
According to Pythagoras theorem in △ABC we get.
AC2 = AB2+BC2
AC2 = (8k)2+(7k)2
AC2 = 64k2+49k2
AC2 = 113k2
AC = √113 k
According to the sine and cos function ratios,it is written as
sin θ = AB/AC = Opposite Side/Hypotenuse =8k/√113 k = 8/√113 and
cos θ = Adjacent Side/Hypotenuse = BC/AC =7k/√113 k = 7/√113
Now apply the values of sin function and cosfunction:
If 3 cot A = 4,check whether (1-tan2 A)/(1+tan2 A) = cos2 A– sin 2 A or not.
Answer - 8 : -
Let △ABC inwhich ∠B=90°
We know that, cot function is the reciprocalof tan function and it is written as
cot(A) = AB/BC = 4/3
Let AB = 4k an BC =3k, where k is a positivereal number.
According to the Pythagorean theorem,
AC2=AB2+BC2
AC2=(4k)2+(3k)2
AC2=16k2+9k2
AC2=25k2
AC=5k
Now, apply the values corresponding to theratios
tan(A) = BC/AB = 3/4
sin (A) = BC/AC = 3/5
cos (A) = AB/AC = 4/5
Now compare the left hand side(LHS) with righthand side(RHS)
Since, both the LHS and RHS = 7/25
R.H.S. =L.H.S.
Hence, (1-tan2 A)/(1+tan2 A) = cos2 A– sin 2 A is proved
Question - 9 : - In triangle ABC,right-angled at B, if tan A = 1/√3 find the value of:
Answer - 9 : -
(i) sin A cos C + cos A sin C
(ii) cos A cos C – sin A sin C
Solution
Question - 10 : - In ∆ PQR,right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sinP, cos P and tan P
Answer - 10 : - 