Exercise 8.1 answers: Introduction to Trigonometry

Open NCERT p. 121Matches NCERT’s answer

(i) sin A, cos A

1. The right angle is at B, so AC is the hypotenuse.
2. By Pythagoras, AC² = AB² + BC² = 24² + 7² = 576 + 49 = 625, so AC = 25 cm.
3. For angle A: the side facing A is BC = 7, and the hypotenuse is AC = 25.
4. sin A = BC/AC = 7/25. The side next to A is AB = 24, so cos A = AB/AC = 24/25.

Answersin A = 7/25, cos A = 24/25

(ii) sin C, cos C

1. For angle C: the side facing C is AB = 24, and the hypotenuse is still AC = 25.
2. sin C = AB/AC = 24/25.
3. The side next to C is BC = 7, so cos C = BC/AC = 7/25.

Answersin C = 24/25, cos C = 7/25

Watch this explained “All six on one triangle”, 8:48 into Defining sine, cosine and tangent, then their three reciprocals

Open NCERT p. 121Matches NCERT’s answer

1. In the figure, the right angle is at Q, PQ = 12 cm and the hypotenuse PR = 13 cm.
2. By Pythagoras, QR² = PR² − PQ² = 169 − 144 = 25, so QR = 5 cm.
3. At P: the facing side is QR = 5 and the near side is PQ = 12, so tan P = 5/12.
4. At R: the side next to R is QR = 5 and the facing side is PQ = 12, so cot R = QR/PQ = 5/12.
5. tan P − cot R = 5/12 − 5/12 = 0.

Answertan P − cot R = 0

Watch this explained “The swap doing real work”, 11:06 into Defining sine, cosine and tangent, then their three reciprocals

Open NCERT p. 121Matches NCERT’s answer

1. sin A = 3/4 means the facing side and the hypotenuse are in the ratio 3 to 4, not that they equal 3 and 4.
2. Write the facing side as 3k and the hypotenuse as 4k, for some positive k.
3. By Pythagoras, the near side² = (4k)² − (3k)² = 16k² − 9k² = 7k², so the near side = √7 k.
4. cos A = near side / hypotenuse = √7k / 4k = √7/4. The k cancels.
5. tan A = facing side / near side = 3k / √7k = 3/√7.

Answercos A = √7/4, tan A = 3/√7

Watch this explained “The multiplier, and what it may be”, 1:35 into Given one ratio, reconstructing the other five

Open NCERT p. 121Matches NCERT’s answer

1. 15 cot A = 8 means cot A = 8/15, so the near side to A and the facing side are in the ratio 8 to 15.
2. Write the near side as 8k and the facing side as 15k.
3. By Pythagoras, hypotenuse² = (8k)² + (15k)² = 64k² + 225k² = 289k², so the hypotenuse = 17k.
4. sin A = facing side / hypotenuse = 15k/17k = 15/17.
5. sec A = hypotenuse / near side = 17k/8k = 17/8.

Answersin A = 15/17, sec A = 17/8

Watch this explained “The same loop, a different start”, 5:10 into Given one ratio, reconstructing the other five

Open NCERT p. 121Matches NCERT’s answer

1. sec θ = 13/12 means hypotenuse/near side = 13/12, so write hypotenuse = 13k, near side = 12k.
2. By Pythagoras, facing side² = (13k)² − (12k)² = 169k² − 144k² = 25k², so facing side = 5k.
3. cos θ = 12k/13k = 12/13. sin θ = 5k/13k = 5/13.
4. tan θ = sin θ/cos θ = 5/12. cosec θ = 1/sin θ = 13/5. cot θ = 1/tan θ = 12/5.

Answersin θ = 5/13, cos θ = 12/13, tan θ = 5/12, cosec θ = 13/5, cot θ = 12/5

Watch this explained “The loop, ready to run again”, 11:48 into Given one ratio, reconstructing the other five

Open NCERT p. 121One way to think about it

1. Put ∠A in a right triangle: let the side next to A be x₁, the side facing A be y₁ and the hypotenuse h₁. Put ∠B in another right triangle with side next to B x₂, side facing B y₂ and hypotenuse h₂.
2. cos A = x₁/h₁ and cos B = x₂/h₂. Since cos A = cos B, call this common value c. Then x₁ = c h₁ and x₂ = c h₂.
3. The hypotenuse is the longest side, so c is less than 1 and 1 − c² is not 0.
4. By Pythagoras, y₁ = √(h₁² − x₁²) = √(h₁² − c²h₁²) = h₁√(1 − c²). In the same way, y₂ = h₂√(1 − c²).
5. So x₁/x₂ = c h₁/(c h₂) = h₁/h₂ and y₁/y₂ = h₁√(1 − c²)/(h₂√(1 − c²)) = h₁/h₂. All three pairs of sides are in the same ratio h₁ : h₂.
6. By the SSS similarity criterion the two triangles are similar, with the corner at A matching the corner at B (both lie between the side next to the angle and the hypotenuse).
7. Corresponding angles of similar triangles are equal, so ∠A = ∠B.

In short∠A = ∠B: equal cosines put all three pairs of sides of the two right triangles in the same ratio, so the triangles are similar and the matching angles A and B are equal.

Watch this explained “Running it backwards”, 11:13 into Why enlarging the triangle leaves every ratio unchanged

Open NCERT p. 121Matches NCERT’s answer

(i) (1 + sin θ)(1 – sin θ)/((1 + cos θ)(1 – cos θ))

1. cot θ = 7/8 means the side next to θ and the side facing θ are in the ratio 7 : 8, so write them as 7k and 8k (k > 0).
2. By Pythagoras, hypotenuse² = (7k)² + (8k)² = 49k² + 64k² = 113k², so the hypotenuse = √113 k.
3. So sin θ = 8k/(√113 k) = 8/√113 and cos θ = 7k/(√113 k) = 7/√113.
4. Top: (1 + sin θ)(1 − sin θ) = 1 − sin²θ = 1 − 64/113 = 49/113.
5. Bottom: (1 + cos θ)(1 − cos θ) = 1 − cos²θ = 1 − 49/113 = 64/113.
6. So the expression = (49/113) ÷ (64/113) = 49/64.

Answer49/64

(ii) cot² θ

1. cot² θ = (7/8)² = 49/64.

Answer49/64

Watch this explained “The same loop, a different start”, 5:10 into Given one ratio, reconstructing the other five

Open NCERT p. 121Matches NCERT’s answer

1. 3 cot A = 4 gives cot A = 4/3, so tan A = 3/4.
2. (1 − tan²A)/(1 + tan²A) = (1 − 9/16)/(1 + 9/16) = (7/16)/(25/16) = 7/25.
3. For cos A and sin A: cot A = 4/3 means near side = 4k, facing side = 3k, so hypotenuse = 5k.
4. cos A = 4/5, sin A = 3/5, so cos²A − sin²A = 16/25 − 9/25 = 7/25.
5. Both sides equal 7/25, so the two expressions are equal here.

AnswerYes, both sides equal 7/25.

Watch this explained “The same loop, a different start”, 5:10 into Given one ratio, reconstructing the other five

Open NCERT p. 121Matches NCERT’s answer

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

1. tan A = BC/AB (BC faces A, AB is next to A) = 1/√3, so write BC = k and AB = √3 k.
2. By Pythagoras, AC² = AB² + BC² = 3k² + k² = 4k², so AC = 2k.
3. At A: sin A = BC/AC = 1/2 and cos A = AB/AC = √3/2.
4. At C the two legs swap roles (AB faces C, BC is next to C): sin C = AB/AC = √3/2 and cos C = BC/AC = 1/2.
5. sin A cos C + cos A sin C = (1/2)(1/2) + (√3/2)(√3/2) = 1/4 + 3/4 = 1.

Answer1

(ii) cos A cos C – sin A sin C

1. With the same values: cos A cos C − sin A sin C = (√3/2)(1/2) − (1/2)(√3/2) = √3/4 − √3/4 = 0.

Answer0

Watch this explained “The same picture, the other end”, 7:55 into Squeezing 30°, 45° and 60° out of two special triangles

Open NCERT p. 121Matches NCERT’s answer

1. Let QR = x, so PR = 25 − x, and PQ = 5.
2. By Pythagoras (right angle at Q), PR² = PQ² + QR²: (25−x)² = 25 + x².
3. Expand: 625 − 50x + x² = 25 + x², so 600 = 50x, giving x = 12.
4. So QR = 12 cm and PR = 25 − 12 = 13 cm.
5. At P: facing side = QR = 12, near side = PQ = 5, hypotenuse = PR = 13.
6. sin P = 12/13, cos P = 5/13, tan P = 12/5.

Answersin P = 12/13, cos P = 5/13, tan P = 12/5

Watch this explained “When the data is not a ratio at all”, 9:26 into Given one ratio, reconstructing the other five

Open NCERT p. 121Matches NCERT’s answer

(i) The value of tan A is always less than 1.

1. tan A = side facing A ÷ side next to A, and nothing stops the side facing A from being the longer leg.
2. For example, in a right triangle with legs 3 cm (next to A) and 4 cm (facing A), tan A = 4/3, which is more than 1.
3. So tan A is not always less than 1.

AnswerFalse

(ii) sec A = 12/5 for some value of angle A.

1. sec A = hypotenuse ÷ side next to A = 12/5, so try a hypotenuse of 12k and a side next to A of 5k.
2. By Pythagoras, (side facing A)² = (12k)² − (5k)² = 144k² − 25k² = 119k², so that side is √119 k, a real positive length.
3. So this right triangle exists, and its angle A has sec A = 12/5.

AnswerTrue

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

1. cos A is short for the cosine of angle A. The cosecant of angle A is written cosec A.

AnswerFalse

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

1. cot A is one symbol meaning the cotangent of angle A. It is not "cot" multiplied by A; "cot" on its own has no value.

AnswerFalse

(v) sin θ = 4/3 for some angle θ.

1. sin θ = side facing θ ÷ hypotenuse.
2. The hypotenuse is the longest side of a right triangle, so this fraction is always less than 1.
3. 4/3 is more than 1, so no angle θ has sin θ = 4/3.

AnswerFalse

Watch this explained “Which values are impossible”, 10:38 into Given one ratio, reconstructing the other five