Mathematics · Introduction to Trigonometry · NCERT Exercises

Exercise 8.1

Complete, independently verified solutions for NCERT Exercise 8.1.

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Mathematics · Chapter 8 · NCERT Exercises

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Verified NCERT questions and solutionsSolutions are checked independently and follow the supplied NCERT chapter edition.

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Question

Question 1

trigonometric ratios · right triangle

In right triangle ABC, right-angled at B, AB = 24 cm and BC = 7 cm. Find (i) sin A, cos A and (ii) sin C, cos C.

Right triangle for Question 1Triangle ABC is right-angled at B with AB = 24 cm, BC = 7 cm and hypotenuse AC = 25 cm.24 cm7 cm25 cmAABC
Right triangle for Question 1 — explanatory sketch (not to scale).

Solution

Board-exam working:

AC=AB2+BC2=242+72=25 cmAC=\sqrt{AB^2+BC^2}=\sqrt{24^2+7^2}=25\text{ cm}

(i) For angle A:

sin⁡A=BCAC=725\sin A=\frac{BC}{AC}=\frac7{25}
cos⁡A=ABAC=2425\cos A=\frac{AB}{AC}=\frac{24}{25}

(ii) For angle C:

sin⁡C=ABAC=2425\sin C=\frac{AB}{AC}=\frac{24}{25}
cos⁡C=BCAC=725\cos C=\frac{BC}{AC}=\frac7{25}

Final answer

(i) sin⁡A=725, cos⁡A=2425;(ii) sin⁡C=2425, cos⁡C=725\text{(i) }\sin A=\frac7{25},\ \cos A=\frac{24}{25};\qquad \text{(ii) }\sin C=\frac{24}{25},\ \cos C=\frac7{25}

Question

Question 2

tan · cot · figure

In Figure 8.13, find tan P − cot R.

Right triangle PQR, right-angled at Q, with PQ 12 centimetres and PR 13 centimetres.
Figure 8-13 from the supplied NCERT chapter.

Solution

Board-exam working:

QR=PR2−PQ2=132−122=5 cmQR=\sqrt{PR^2-PQ^2}=\sqrt{13^2-12^2}=5\text{ cm}
tan⁡P=QRPQ=512\tan P=\frac{QR}{PQ}=\frac5{12}
cot⁡R=QRPQ=512\cot R=\frac{QR}{PQ}=\frac5{12}
tan⁡P−cot⁡R=512−512=0\tan P-\cot R=\frac5{12}-\frac5{12}=0

Final answer

0

Question

Question 3

given sine · Pythagoras theorem

If sin A = 3/4, calculate cos A and tan A.

Sine ratio with opposite side 3An acute angle A in a right triangle has opposite side 3k, hypotenuse 4k and adjacent side square root of 7 times k.√7 k3k4kAABC
Sine ratio with opposite side 3 — explanatory sketch (not to scale).

Solution

Board-exam working:

sin⁡A=oppositehypotenuse=34\sin A=\frac{\text{opposite}}{\text{hypotenuse}}=\frac34

Let opposite side = 3k and hypotenuse = 4k.

adjacent side=(4k)2−(3k)2=k7\text{adjacent side}=\sqrt{(4k)^2-(3k)^2}=k\sqrt7
cos⁡A=74\cos A=\frac{\sqrt7}{4}
tan⁡A=37=377\tan A=\frac3{\sqrt7}=\frac{3\sqrt7}{7}

Final answer

cos⁡A=74,tan⁡A=377\cos A=\frac{\sqrt7}{4},\qquad \tan A=\frac{3\sqrt7}{7}

Question

Question 4

cotangent · right triangle

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

Cotangent ratio 8 to 15An acute angle A in an 8-15-17 right triangle has adjacent side 8k and opposite side 15k.8k15k17kAABC
Cotangent ratio 8 to 15 — explanatory sketch (not to scale).

Solution

Board-exam working:

15cot⁡A=8⇒cot⁡A=81515\cot A=8\Rightarrow \cot A=\frac8{15}

Let adjacent side = 8k and opposite side = 15k.

hypotenuse=(8k)2+(15k)2=17k\text{hypotenuse}=\sqrt{(8k)^2+(15k)^2}=17k
sin⁡A=15k17k=1517\sin A=\frac{15k}{17k}=\frac{15}{17}
sec⁡A=17k8k=178\sec A=\frac{17k}{8k}=\frac{17}{8}

Final answer

sin⁡A=1517,sec⁡A=178\sin A=\frac{15}{17},\qquad \sec A=\frac{17}{8}

Question

Question 5

secant · all ratios

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

Secant ratio 13 to 12An acute angle theta in a 5-12-13 right triangle has adjacent side 12k and hypotenuse 13k.12k5k13kθABC
Secant ratio 13 to 12 — explanatory sketch (not to scale).

Solution

Board-exam working:

sec⁡θ=hypotenuseadjacent=1312\sec\theta=\frac{\text{hypotenuse}}{\text{adjacent}}=\frac{13}{12}

Let hypotenuse = 13k and adjacent side = 12k.

opposite side=(13k)2−(12k)2=5k\text{opposite side}=\sqrt{(13k)^2-(12k)^2}=5k
sin⁡θ=513\sin\theta=\frac5{13}
cos⁡θ=1213\cos\theta=\frac{12}{13}
tan⁡θ=512\tan\theta=\frac5{12}
cot⁡θ=125\cot\theta=\frac{12}{5}
cosec⁡θ=135\cosec\theta=\frac{13}{5}

Final answer

sin⁡θ=513, cos⁡θ=1213, tan⁡θ=512, cot⁡θ=125, cosec⁡θ=135\sin\theta=\frac5{13},\ \cos\theta=\frac{12}{13},\ \tan\theta=\frac5{12},\ \cot\theta=\frac{12}{5},\ \cosec\theta=\frac{13}{5}

Question

Question 6

equal cosines · acute angles

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

Comparing two equal cosine ratiosRight triangles ACD and BEF have acute angles A and B at their top vertices. C and E are the right angles, so neither acute angle is confused with the right-angle vertex.adjacenthypotenuseadjacenthypotenuseABACDBEF
Comparing two equal cosine ratios — explanatory sketch (not to scale).

Solution

Board-exam working:

For acute angles, cosine has one distinct value for each angle and decreases from 1 to 0 as the angle increases from 0° to 90°.

cos⁡A=cos⁡B\cos A=\cos B

Therefore A and B cannot be different acute angles.

∠A=∠B\angle A=\angle B

Final answer

Hence, ∠A = ∠B.

Question

Question 7

cotangent · identity

If cot θ = 7/8, evaluate (i) [(1 + sin θ)(1 − sin θ)]/[(1 + cos θ)(1 − cos θ)] and (ii) cot²θ.

Cotangent ratio 7 to 8An acute angle theta has adjacent side 7k, opposite side 8k and hypotenuse square root of 113 times k.7k8k√113 kθABC
Cotangent ratio 7 to 8 — explanatory sketch (not to scale).

Solution

Board-exam working:

(i)

(1+sin⁡θ)(1−sin⁡θ)(1+cos⁡θ)(1−cos⁡θ)=1−sin⁡2θ1−cos⁡2θ\frac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}=\frac{1-\sin^2\theta}{1-\cos^2\theta}
=cos⁡2θsin⁡2θ=cot⁡2θ=\frac{\cos^2\theta}{\sin^2\theta}=\cot^2\theta
=(78)2=4964=\left(\frac78\right)^2=\frac{49}{64}

(ii)

cot⁡2θ=(78)2=4964\cot^2\theta=\left(\frac78\right)^2=\frac{49}{64}

Final answer

(i) 4964;(ii) 4964\text{(i) }\frac{49}{64};\qquad \text{(ii) }\frac{49}{64}

Question

Question 8

verification · identity

If 3 cot A = 4, check whether (1 − tan²A)/(1 + tan²A) = cos²A − sin²A.

Cotangent ratio 4 to 3An acute angle A in a 3-4-5 right triangle has adjacent side 4k and opposite side 3k.4k3k5kAABC
Cotangent ratio 4 to 3 — explanatory sketch (not to scale).

Solution

Board-exam working:

3cot⁡A=4⇒cot⁡A=43⇒tan⁡A=343\cot A=4\Rightarrow \cot A=\frac43\Rightarrow \tan A=\frac34

Left-hand side:

LHS=1−(34)21+(34)2=725\mathrm{LHS}=\frac{1-\left(\frac34\right)^2}{1+\left(\frac34\right)^2}=\frac{7}{25}

Using sides 3, 4 and 5:

sin⁡A=35,cos⁡A=45\sin A=\frac35,\qquad \cos A=\frac45

Right-hand side:

RHS=(45)2−(35)2=725\mathrm{RHS}=\left(\frac45\right)^2-\left(\frac35\right)^2=\frac7{25}
LHS=RHS\mathrm{LHS}=\mathrm{RHS}

Final answer

Yes, the equality is true.

Question

Question 9

complementary angles · special angles

In triangle ABC, right-angled at B, tan A = 1/√3. Find (i) sin A cos C + cos A sin C and (ii) cos A cos C − sin A sin C.

Complementary acute angles A and CTriangle ABC is right-angled at B. Angle A is 30 degrees and angle C is 60 degrees; adjacent side AB is square root of 3, opposite side BC is 1 and hypotenuse AC is 2.√31230°60°ABC
Complementary acute angles A and C — explanatory sketch (not to scale).

Solution

Board-exam working:

tan⁡A=13⇒A=30∘\tan A=\frac1{\sqrt3}\Rightarrow A=30^\circ
C=90∘−A=60∘C=90^\circ-A=60^\circ

(i)

sin⁡30∘cos⁡60∘+cos⁡30∘sin⁡60∘\sin30^\circ\cos60^\circ+\cos30^\circ\sin60^\circ
=12⋅12+32⋅32=14+34=1=\frac12\cdot\frac12+\frac{\sqrt3}{2}\cdot\frac{\sqrt3}{2}=\frac14+\frac34=1

(ii)

cos⁡30∘cos⁡60∘−sin⁡30∘sin⁡60∘\cos30^\circ\cos60^\circ-\sin30^\circ\sin60^\circ
=32⋅12−12⋅32=0=\frac{\sqrt3}{2}\cdot\frac12-\frac12\cdot\frac{\sqrt3}{2}=0

Final answer

(i) 1; (ii) 0.

Question

Question 10

right triangle · trigonometric ratios

In right triangle PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Find sin P, cos P and tan P.

Right triangle PQR for Question 10Triangle PQR is right-angled at Q with PQ = 5 cm, QR = 12 cm and PR = 13 cm.5 cm12 cm13 cmPPQR
Right triangle PQR for Question 10 — explanatory sketch (not to scale).

Solution

Board-exam working:

Let QR = x cm. Then PR = 25 − x cm.

PR2=PQ2+QR2PR^2=PQ^2+QR^2
(25−x)2=52+x2(25-x)^2=5^2+x^2
625−50x+x2=25+x2625-50x+x^2=25+x^2
x=12x=12
QR=12 cm,PR=13 cmQR=12\text{ cm},\qquad PR=13\text{ cm}
sin⁡P=1213\sin P=\frac{12}{13}
cos⁡P=513\cos P=\frac5{13}
tan⁡P=125\tan P=\frac{12}{5}

Final answer

sin⁡P=1213,cos⁡P=513,tan⁡P=125\sin P=\frac{12}{13},\qquad \cos P=\frac5{13},\qquad \tan P=\frac{12}{5}

Question

Question 11

true or false · ratio definitions

State true or false and justify: (i) tan A is always less than 1; (ii) sec A = 12/5 for some angle A; (iii) cos A abbreviates cosecant A; (iv) cot A is the product of cot and A; (v) sin θ = 4/3 for some angle θ.

Ratio definitions for true-or-false statementsA right triangle marks angle A, its opposite side, adjacent side and hypotenuse so the trigonometric ratios can be read directly.adjacentoppositehypotenuseAABC
Ratio definitions for true-or-false statements — explanatory sketch (not to scale).

Solution

Board-exam working:

(i) False.

tan⁡45∘=1andtan⁡60∘=3>1\tan45^\circ=1\quad\text{and}\quad\tan60^\circ=\sqrt3>1

(ii) True.

sec⁡A=125⇒cos⁡A=512\sec A=\frac{12}{5}\Rightarrow \cos A=\frac5{12}

Since 0 < 5/12 < 1, such an acute angle exists.

(iii) False. cos A means cosine of A; cosecant is written cosec A.

(iv) False. cot A denotes the cotangent ratio of angle A.

(v) False.

0≤sin⁡θ≤1,but43>10\leq\sin\theta\leq1,\quad\text{but}\quad\frac43>1

Final answer

(i) False; (ii) True; (iii) False; (iv) False; (v) False.

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