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Question 1:

Given $sec\Theta =\frac{13}{12}$ . Calculate all other trigonometric ratios.

Solution 1:

Consider a right-angle triangle ΔABC, right-angled at point B.

$sec\theta =\frac{hypotenuse}{side adjacent to\angle \theta }$

$=\frac{13}{12}=\frac{AC}{AB}$

If AC is 13k, AB will be 12k, where k is a positive integer.

Applying Pythagoras theorem in ΔABC, we obtain

(AC)² = (AB)² + (BC)²

(13k)² = (12k)² + (BC)²

169k² = 144k² + BC²

25k² = BC²

BC = 5k

$sin\theta =\frac{side opposite to \angle \theta }{hypotenuse}$

$=\frac{BC}{AC}=\frac{5k}{13k}$ $=\frac{5}{13}$

$cos\theta =\frac{side adjacent}{hypotenuse}=\frac{AB}{AC}=\frac{12k}{13k}=\frac{12}{13}$

$tan\theta =\frac{side opposite}{side adjacent}=\frac{BC}{AB}=\frac{5k}{12k}=\frac{5}{12}$

$cot\theta =\frac{side adjacent to \angle \theta }{side opposite to\angle \theta }=\frac{AB}{BC}=\frac{12K}{5k}=\frac{12}{5}$

$cosec\theta =\frac{hypotenuse}{side opposite}=\frac{AC}{BC}=\frac{13K}{5k}=\frac{13}{5}$

Comments

Tasneem

Question 2:

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

(i) sin A, cos A (ii) sin C, cos C

Solution 2:

In Δ ABC, right-angled at B

Using Pythagoras theorem

AC² = AB² +BC²

AC² = 576 + 49 = 625

AC = √625

AC = ±25

Now

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

Sin A = BC/AC

=7/25

CosA =AB/AC

=24/25

(ii) Sin C =AB/AC

=24/25

Cos C =BC/AC

=7/25

Comments

Shubhangi B.

Question 3:

In the given figure, find tan P – cot R.

Solution 3:

Using pythagoras theorem,

(PR) ² = (PQ)² + (QR)²

⇒(QR)² = (13)² - (12) =169 – 144 = 25;

QR = 5 cm

tan p =(opposite side ÷adjacentside) = $\frac{QR}{PQ}$ = $\frac{5}{12}$

cot R= (adjacent side÷ opposite side) = $\frac{QR}{PQ}$ = $\frac{5}{12}$

tan P- cot R= $\frac{5}{12}-\frac{5}{12}=0$

Comments

Sherine Das

Question 4:

If sin A = $\frac{3}{4}$ . Calculate cos A and tan A.

Solution 4:

$CosA = \frac{\sqrt{7}}{4}$

$tanA=\frac{3}{\sqrt{7}}$

Step-by-step explanation:

Given $sinA = \frac{3}{4}$

i ) $cosA = \sqrt{1-sin^{2}A}$

= $\sqrt{1-\left(\frac{3}{4}}\right)^{2}$

/* From (1) */

= $\sqrt{1-\frac{9}{16}}$

= $\sqrt{\frac{16-9}{16}}$

= $\sqrt{\frac{7}{16}}$

= $\frac{\sqrt{7}}{4}$ ---(1)

ii) $tanA = \frac{SinA}{cosA}$

= $\frac{\frac{3}{4}}{\frac{\sqrt{7}}{4}}$

After cancellation, we get

= $\frac{3}{\sqrt{7}}$ ---(2)

Therefore,

$CosA = \frac{\sqrt{7}}{4}$

$tanA=\frac{3}{\sqrt{7}}$

Comments

Shubhangi B.

Question 5:

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

Solution 5:

Consider a right-angled triangle, right-angled at B.

$cotA=\frac{side adjacent to \angle A}{side opposite to\angle A}$

= $\frac{AB}{BC}$

It is given that,

cot A =

Let AB be 8k.Therefore, BC will be 15k, where k is a positive integer.

Applying Pythagoras theorem in ΔABC, we obtain

AC² = AB² + BC²

= (8k)² + (15k)²

= 64k² + 225k²

= 289k²

AC = 17k

$sinA=\frac{side opposite to \angle A}{hypotenuse}$

$=\frac{BC}{AC}$ $=\frac{15k}{17k}=\frac{15}{17}$

$secA=\frac{hypotenuse}{side adjacent to\angle A}$

$\frac{AC}{AB}=\frac{17}{8}$

Comments

Tasneem

Question 6:

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.

Solution 6:

Given PR + QR = 25 , PQ = 5

PR be x. and QR = 25 - x

Applying Pythagoras theorem in ΔPQR, we obtain

PR² = PQ² + QR²

x² = (5)² + (25 − x)²

x² = 25 + 625 + x² − 50x

50x = 650

x = 13

Therefore, PR = 13 cm

QR = (25 − 13) cm = 12 cm

$sin P=\frac{QR}{PR}=\frac{12}{13}$

$cosP= \frac{PQ}{PR}=\frac{5}{13}$

$tanP= \frac{QR}{PQ}=\frac{12}{5}$

Comments

Reetika

Question 7:

If $\angle A$ and $\angle B$ are acute angles such that cos A = cos B, then show that $\angle A$ and $\angle B$ .

Solution 7:

Let us consider a triangle ABC in which CD ⊥ AB.

It is given that

cos A = cos B

… (1)

We have to prove ∠A = ∠B. To prove this, let us extend AC to P such that BC = CP.

From equation (1), we get

$\frac{AD}{BD}=\frac{AC}{BC}$

$\Rightarrow \frac{AD}{BD}=\frac{AC}{CP}$ (through construction we have BC=CP) .............(2)

By using the converse of B.P.T,

CD||BP

⇒∠ACD = ∠CPB (Corresponding angles) … (3)

And, ∠BCD = ∠CBP (Alternate interior angles) … (4)

By construction, we have BC = CP.

∴ ∠CBP = ∠CPB (Angle opposite to equal sides of a triangle) … (5)

From equations (3), (4), and (5), we obtain

∠ACD = ∠BCD … (6)

In ΔCAD and ΔCBD,

∠ACD = ∠BCD [Using equation (6)]

∠CDA = ∠CDB [Both 90°]

Therefore, the remaining angles should be equal.

∴∠CAD = ∠CBD

⇒ ∠A = ∠B

Comments

Tasneem

Question 8:

If $cot\Theta =\frac{7}{8}$ evaluate:

$(i) \frac{(1+sin\Theta )(1-sin\Theta )}{(1+cos\Theta)(1-cos\Theta )}$

(ii) $cot^{2}\Theta$

Solution 8:

Let us consider a right triangle ABC, right-angled at point B.

$cot\theta =\frac{side adjacent to \angle \theta }{side opposite to\angle \theta }=\frac{BC}{AB}=\frac{7}{8}$

If BC is 7k, then AB will be 8k, where k is a positive integer.

Applying Pythagoras theorem in ΔABC, we obtain

AC² = AB² + BC²

= (8k)² + (7k)²

= 64k² + 49k²

= 113k²

AC =

$sin\theta =\frac{side opposite to \angle \theta }{hypotenuse} =\frac{AB}{AC}=\frac{8k}{\sqrt{113}k}=\frac{8}{\sqrt{113}}$

$cos\theta =\frac{side adjacent}{hypotenuse}=\frac{BC}{AC}=\frac{7k}{\sqrt{113}k}=\frac{7}{\sqrt{113}}$

(i) $\frac{(1-sin \theta)(1+sin \theta)}{(1-cos \theta)(1+cos \theta)} = \frac{1- sin^2 \theta}{1-cos^2 \theta} = \frac{cos^2 \theta}{sin^2 \theta} =\frac{49}{64}$

(ii) cot² θ = (cot θ)² = =

Comments

Tasneem

Question 9:

If 3 cot A = 4, check whether $\frac{1-tan^{2}A}{1+tan^{2}A}=cos^{2}A-sin^{2}A$ or not.

Solution 9:

Given

Cot A = 4/3 Or B/P=4/3

Let B=4k and P=4k

So in a right angle triangle with angle A

$P^{2} + B^{2} = H^{2}$ (where P=perpendicular, B=Base, H=Hypotenuse)

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

$[1-tan^{2}A]/ [1+tan^{2}A]$ = $[1-(3/4)^2]/[1+(3/4)^{2}] = 7/25$

RHS = $cos^{2} A - sin^{2} A = 7/25$

So LHS=RHS, sothe statement in true

Comments

Reetika

Question 10:

In triangle ABC, right-angled at B, if $tanA=\frac{1}{\sqrt{3}}$ find the value of:

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

(ii) cos A cos C – sin A sin C

Solution 10:

If BC is k, then AB will be, where k is a positive integer.

In ΔABC,

AC² = AB² + BC²

= 3k² + k² = 4k²

∴ AC = 2k

$sinA=\frac{BC}{AC}=\frac{k}{2k}=\frac{1}{2}$

$cosA=\frac{AB}{AC}=\frac{\sqrt{3}k}{2k}=\frac{\sqrt{3}}{2}$

$sinC=\frac{AB}{AC}=\frac{\sqrt{3}k}{2k}=\frac{\sqrt{3}}{2}$

$cosC=\frac{BC}{AC}=\frac{k}{2k}=\frac{1}{2}$

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

$\left ( \frac{1}{2} \right )\left ( \frac{1}{2} \right )+\left ( \frac{\sqrt{3}}{2} \right )\left ( \frac{\sqrt{3}}{2} \right )$

$=\frac{1}{4}+\frac{3}{4}=1$

(ii) cos A cos C − sin A sin C

= $\left ( \frac{\sqrt{3}}{2} \right )\left ( \frac{1}{2} \right )-\left ( \frac{1}{2} \right )\left ( \frac{\sqrt{3}}{2} \right )$

$=\frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{4}=0$

Comments

Tasneem

Question 11:

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 = $\frac{12}{5}$ for some value of angle A.

(iii) cos A is the abbreviation used for the cosecant of angle A.

(iv) cot A is the product of cot and A.

(v) $sin\Theta =\frac{4}{3}$ for some angle $\Theta$ .

Solution 11:

(i) Consider a ΔABC, right-angled at B.

$tanA= \frac{side opposite}{side adjacent }=\frac{12}{5}$

but $\frac{12}{5}> 1$

$\therefore tanA> 1$

Hence, the given statement is false.

(ii) $secA=\frac{12}{5}$

$secA=\frac{AC}{AB}=\frac{12}{5}$

Let AC be 12k, AB will be 5k, where k is a positive integer.

Applying Pythagoras theorem in ΔABC, we obtain

AC² = AB² + BC²

(12k)² = (5k)² + BC²

144k² = 25k² + BC²

BC² = 119k²

BC = 10.9k

It can be observed that for given two sides AC = 12k and AB = 5k,

BC should be such that,

AC − AB < BC < AC + AB

12k − 5k < BC < 12k + 5k

7k < BC < 17 k

However, BC = 10.9k. Clearly, such a triangle is possible and hence, such value of sec A is possible.

Hence, the given statement is true.

(iii) Abbreviation used for cosecant of angle A is cosec A. And cos A is the abbreviation used for cosine of angle A.

Hence, the given statement is false.

(iv) cot A is not the product of cot and A. It is the cotangent of ∠A.

Hence, the given statement is false.

$(v)sin\theta =\frac{4}{3}$

$sin\theta =\frac{side opposite }{hypotenuse}$

In a right-angled triangle, hypotenuse is always greater than the remaining two sides. Therefore, such value of sin θ is not possible.

Hence, the given statement is false.

Comments

Tasneem

All Exercises - Chapter 8 - Introduction to Trigonometry

Exercise 8.1

Exercise 8.2

Exercise 8.3

Exercise 8.4

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All Exercises - Chapter 8 - Introduction to Trigonometry

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