10th Maths 8.1
Chapter 8
Introduction to Trigonometry
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NCERT Class 10th solution of Exercise 8.2
NCERT Class 10th solution of Exercise 8.3
NCERT Class 10th solution of Exercise 8.4
Exercise 8.1
Q1. In `triangle ABC`, right-angled at `B`, `AB=24 cm, BC=7 cm`. Determine:
i) sin `A`, cos `A`
ii) sin `C`, cos `C`.
Sol. :
By Pythagoras theorem
`AC^2 = AB^2 + BC^2`
`AC^2 = (24)^2+(7)^2`
`AC^2 = 576 + 49`
`AC^2 = 625`
`AC^2 = (25)^2`
`AC = 25 cm`
i)
`sin A = (BC)/(AC) = 7/25`
`cos A = (AB)/(AC) = 24/25`
Answer:
`sin A = 7/25, cos = 24/25`
ii)
`sin C = (AC)/(AC) = 24/25`
`cos C = (BC)/(AC) = 7/25`
Answer:
`sin C = 24/25, cos C = 7/25`
Q2. In figure, find `tanP-cotR.`
In `triangle PQR,`
By Pythagoras theorem
`QR^2 = PR^2 - PQ^2`
`QR^2 = (13)^2 - (12)^2`
`QR^2 = 169 - 144`
`QR^2 = 25`
`QR^2 = (5)^2`
`QR = 5 cm`
`= tan P - cot R`
`= (QR)/(PQ) - (QR)/(PQ)`
`= 5/12 - 5/12`
`= 0`
Answer:
`tan P - cot R = 0.`
Q3. If `sin A=3/4,` calculate `cos A` and `tan A.`
Sol. :
`sin A = 3/4`
To Find:
`cos A, tan A`
Solve:
In `triangle ABC, angle B = 90^circ`
`sin A = 3/4 = (BC)/(AC)`
By Pythagoras theorem
`AB^2 = AC^2 - BC^2`
`AB^2 = (4)^2 - (3)^2`
`AB^2 = 16 - 9`
`AB^2 = 7`
`AB = sqrt7`
`cos A = (AB)/(AC) = (sqrt7)/4`
`tan A = (BC)/(AB) = 3/(sqrt7)`
Answer:
`cos A = (sqrt7)/4, tan A = 3/(sqrt7)`
Q4. Given `15 cot A=8,` find `sin A` and `sec A`.
Sol. :
`cot A = 8/15`
To find:
`sin A` and `sec A`.
Solve:
In `triangleABC`
`cot A = 8/15 = (AB)/(BC)`
By Pythagoras theorem
`AC^2 = AB^2 + BC^2`
`AC^2 = (8)^2 + (15)^2`
`AC^2 = 64 + 225`
`AC^2 = 289`
`AC^2 = (17)^2`
`AC = 17`
`sin A = (BC)/(AC) = (15)/(17)`
`sec A = (AC)/(AB) = (17)/8`
Answer:
`sin A = (15)/(17)` and `sec A = (17)/8`.
Q5. Given `sec theta=13/12,` calculate all other trigonometric ratios.
Sol. :
`sec theta = 13/12`
To Find:
All trigonometric ratios.
Solve:
In `triangle ABC`
`sec theta = 13/12 = (AC)/(BC)`
By Pythagoras theorem
`AB^2 = AC^2 - BC^2`
`AB^2 = (13)^2 - (12)^2`
`AB^2 = 169 -144`
`AB^2 = 25`
`AB^2 = (5)^2`
`AB = 5`
`sin theta = (AB)/(AC) = 5/12`
`cos theta = (BC)/(AC) = 12/13`
`tan theta = (AB)/(BC) = 5/12`
`cosec theta = (AC)/(AB) =13/5`
`cot theta = (BC)/(AB) = 12/5`
Answer:
`sin theta = 5/12`
`cos theta = 12/13`
`tan theta = 5/12`
`cosec theta =13/5`
`cot theta = 12/5`
Q6. If `angle A` and `angle B` are acute angles such that `cos A = cos B,` then show that
`angle A=angle B.`
Sol. :
`angle A ` and `angle B` are acute angles.
`cos A = cos B`
To Show:
`cos A = cos B`
Solve:
Let `triangle ABC`
`cos A = cos B`
`(AC)/(AB) = (BC)/(AB)`
`AC = BC`
`angle A = angle B` [Angle opposite to equal sides are equal.]
Proved.
Q7. If `cot theta=7/8,` evaluate:
i) `((1+sin theta)(1-sin theta))/((1+cos theta)(1-cos theta))`
ii) `cot^2 theta`
Sol. :
i)
`= ((1 + sin theta)(1 - sin theta))/((1 + cos theta)(1 - cos theta))` [`(a+b)(a-b)=a^2-b^2`]
`= (1 - sin^2 theta)/(1 - cos^2 theta)`
`= (cos^2 theta)/(sin^2 theta)`
`= (cot theta)^2`
`= cot^2 theta`
`= (7/8)^2`
`=49/64`
Answer:
`=49/64`
ii)
`= cot^2 theta`
`= (cot theta)^2`
`= (7/8)^2`
`= 49/64`
Answer:
`= 49/64`
Q8. If `3cot A=4,` check whether `(1-tan^2 A)/(1+tan^2 A)=cos^2 A-sin^2 A` or not.
Sol. :
`cot A = 4/3`
Solve:
Let `triangle ABC`
`cot A = 4/3 = (AB)/(BC)`
By Pythagoras theorem
`AC^2 = AB^2 + BC^2`
`AC^2 = (4)^2 + (3)^2`
`AC^2 = 16 + 9`
`AC^2 = 25`
`AC^2 = (5)^2`
`AC = 5`
To Prove:
`(1-tan^2 A)/(1+tan^2 A)=cos^2 A-sin^2 A` or not.
Proof:
`LHS.`
`= (1- (BC)/(AB))^2/(1+(BC)/(AB))^2`
`= (1-(3/4)^2)/(1+(3/4)^2)`
`= (1-9/16)/(1+9/16)`
`= (7/16)/(25/16)`
`= 7/25`
`RHS.`
`= cos^2A-sin^2A`
`= ((AB)/(AC))^2-((BC)/(AC))^2`
`= (4/5)^2-(3/5)^2`
`= 16/5-9/25`
`= (16-9)/25`
`= 7/25`
`LHS = RHS`
Proved.
Q9. In triangle `ABC,` right-angled at `B,` if `tan A=1/sqrt3,` find the value of
i) `sin A cos C + cos A sin C`
ii) `cos A cos C - sin A sin C`
Sol. :
`tan A = 1/sqrt3`
Solve:
In `triangle ABC`
`tan A = 1/sqrt3 = (BC)/(AB)`
By Pythagoras theorem
`AC^2 = AB^2 + BC^2`
`AC^2 = (sqrt3)^2 + (1)^2`
`AC^2 = 3 + 1`
`AC^2 = 4`
`AC^2 = (2)^2`
`AC = 2`
i)
`= sin A cos C + cos A + sin C`
`= (BC)/(AC).(BC)/(AC) + (AB)/(AC).(AB)/(AC)`
`= ((BC)/(AC))^2 + ((AB)/(AC))^2`
`= (1/2)^2 + (sqrt3/2)^2`
`= 1/4 + 3/4`
`= 4/4 = 1`
Answer:
The value is `1`.
ii)
`= cos A cos C - sin A sin C`
`= (AB)/(AC).(BC)/(AC) - (BC)/(AC).(AB)/(AC)`
`= (AB.BC)/(AC)^2 - (AB.BC)/(AC)^2`
`= (sqrt3.1)/(2)^2 - (sqrt3.1)/(2)^2`
`= (sqrt3)/4 - (sqrt3)/4`
`= 0`
Answer:
The value is `0`.
Q10. In `triangle PQR,` right-angled at `Q, PR+QR=25 cm` and `PQ=5 cm.` Determine the values of `sin P, cos P` and `tan P.`
Sol. :
`triangle PQR,` `angle Q = 90^circ, `
`PR + QR = 25 cm,`___________(1)
`PQ = 5 cm`.
To find:
`sin P, cos P,` and `tan P`.
Solve:
In `triangle PQR`
By Pythagoras theorem
`(PQ)^2 = (PR)^2 - (QR)^2`
`(5)^2 = (PR + QR)(PR - QR)`
`25 = 25(PR - QR)`
`25/25 = PR - QR`
`PR - QR = 1`____________(2)
From equation (1) `+` (2)
`PR + QR + PR - QR = 25 + 1`
`PR + PR = 26`
`2PR = 26`
`PR = 26/2`
`PR = 13`
Put in equation (1)
`13 + QR = 25`
`QR = 25 - 13`
`QR = 12`
`sin P = (QR)/(PR) = 12/13`
`cos P = (PQ)/(PR) = 5/13`
`tan P = (QR)/(QP) = 12/5`
Answer:
`sin P = 12/13, cos P = 5/13, tan P = 12/5`.
Q11. State whether the following are true or false, Justify your answer:
i) The value of `tan A` is always less than `1.`
ii) `sec A=12/5` for some value of angle `A.`
iii) `cos A` is the abbreviation used for the cosecant of angle `A.`
vi) `cot A` is the product of `cot` and `A.`
v) `sin theta=4/3` for some angle `theta.`
Answer:
i) false, ii) true iii) false iv) false v) false.
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