Question

In a right angle triangle ABC with right angle at B, in which a = 24 units, b = 25 units and ∠BAC = θ. Then, find cos θ and tan θ.

Answer 1

Question 1

In Δ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:

(i) Sin A, cos A

(ii) Sin C, cos C

Solution

In Δ ABC, right-angled at B ,using Pythagoras theorem we have

AC2 = AB2 +BC2 = 576 + 49 = 625

Or AC=25 ( taking positive value only)

Now

(i) In a right angle triangle ABC where B=90° ,

NCERT Solutions for Class 10 Maths Chapter 8: Introduction to Trigonometry

Sin A = BC/AC = 7/25

CosA = AB/AC = 24/25

(ii)

NCERT Solutions for Class 10 Maths Chapter 8: Introduction to Trigonometry

Sin C = AB/AC = 24/25

Cos C = BC/AC =7/25

Question 2

In below find tan P – cot R

NCERT Solutions for Class 10 Maths Chapter 8: Introduction to Trigonometry

Solution

Again consider the above figure . Now by Pythagoras theorem

PQ2 + QR2 =PR2

QR=5

Now

tan P = Perp/Base = 5/12

Cot R = Base/Perm = 5/12

So tan P – cot R=0

3. If sinA=34. Calculate cos A and tan A.

Solution

Given sinA=34

Or P/H=3/4

Let P=3k and H=4k

NCERT Solutions for Class 10 Maths Chapter 8: Introduction to Trigonometry

Now By Pythagoras theorem

P2 + B2 =H2

9k2 + B2 =16k2

Or B=+k7–√

Now CosA=BH=7√4

Now tanA=SinACosA=37√

Question 4

Given 15 cot A = 8, find sin A and sec A.

Solution 4

CotA=815

Or

BP=815

Let B=8K and P=15k

class 10 trigonometry ncert solutions figure

So in a right angle triangle with angle A

P2 + B2 =H2

Or H=17K

Sin A = P/H = 15/17

Sec A = H/B = 17/8

Question 5

Given sec θ =13/12. Calculate all other trigonometric ratios.

Solution 5

Given sec θ=13/12

Or

H/B=13/12

let H=13K ,B=12K

Trigonometry NCERT solution question 5 figure

So in a right angle triangle with angle A

P2 + B2 =H2

P=5k

sin θ = P/H = 5/13

Cos θ = B/H = 12/13

Tan θ = P/B = 5/12

Cosec θ= 1/sin θ = 13/5

cot θ = 1/tan θ = 12/5

Question 6

If ∠ A and ∠ B are acute angles such that cos A = cos B, then show that ∠ A = ∠ B.

Solution 6

In a triangle

Cos A =cos B

NCERT Solutions for Class 10 Maths Chapter 8: Introduction to Trigonometry

AC/AB= BC/AB

⇒AC=BC

⇒Angle A and Angle B

Question 7

If cot θ =7/8 evaluate

(i) (1+sinθ)(1−sinθ)(1+cosθ)(1−cosθ)

(ii) cot2θ

Solution

Given

Given cot θ=7/8

Or

B/P=7/8

let B=7K ,P=8K

Trigonometry NCERT solution question 7 figure

So in a right angle triangle with angle θ

P2 + B2 =H2

H=k113−−−√

sinθ=PH=8113√

cosθ=BH=7113√

(i) (1+sinθ)(1−sinθ)(1+cosθ)(1−cosθ)

=1−sin2θ1−cos2θ

1−641131−49113=4964

(ii) Cot2 θ=(cot θ)2= 49/64

Question 8

If 3 cot A = 4, check whether below is true or not

1−tan2A1+tan2A=cos2A−sin2A

Solution 8

Given

Cot A=4/3

Or

B/P=4/3

Let B=4k and P=3k

NCERT Solutions for Class 10 Maths Chapter 8: Introduction to Trigonometry

So in a right angle triangle with angle A

P2 + B2 =H2

H=5k

Now tan A=1/cot A=3/4

Cos A=B/H=4/5

Sin A=P/H=3/5

Let us take the LHS

1−tan2A1+tan2A=1−(34)21+(34)2=725

RHS=cos2 A - sin2A= 7/25

So LHS=RHS,so the statement is true

Question 9

In triangle ABC, right-angled at B, if tanA=13√

Find the value of:

(i) sin A cos C + cos A sin C

(ii) cos A cos C – sin A sin C

Solution

tanA=13√

PB=13√

Let P=k and B=k3–√

Now by Pythagoras theorem

P2 + B2 =H2

H=2k

(i)

SinACosC+CosASinC=(PH)(BH)+(BH)(PH)

=(BCAC)(BCAC)+(ABAC)(ABAC)

=k24k2+3k24k2=1

(ii)

CosACosC−SinASinC=(PH)(PH)+(BH)(BH)

=(BCAC)(ABAC)−(ABAC)(BCAC)

=0

Question 10

In Δ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of

sin P, cos P and tan P.

Solution

Let QR=x andPR=y

Then x+y=25

y=25-x

Now by Pythagorus theorem

x2 + 25= y2

x2 + 25==(25-x)2

Solving it ,we get

X=12 cm

Then y=25-12=13 cm

Now Sin P= 12/13

Cos P=5/13

Tan P=12/5

Answer 2

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