Introduction to Trigonometry

Solution NCERT Exercise 8.1: part-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:

Let the given triangle ABC is depicted in the figure as follows.

Given, ∠ B = 90^o (i.e. right angle)

AB = 24 cm

BC = 7 cm

(Strategy to solve the question: Here first of all we have to calculate the third side of the given triangle. And then we can calculate the required ratios easily.)

We know that, according to Pythagoras Theorem, in a right-angled triangle

(hypotenuse) ² = (perpendicular) ² + (base) ²

∴ for the given triangle ABC

AC² = BC² + AB²

After substituting the values of BC and AB

= (7 cm) ² + (24 cm)²

= 49 cm² + 576 cm²

⇒ AC² = 625 cm²

⇒ AC = (625 cm)²

⇒ AC = 25 cm = Hypotenuse (h)

(i) sin A, cos A

Solution:

For, acute angle A

Here, we have,

Perpendicular (opposite side of the acute angle A) [p]= BC = 7 cm

Base (adjacent side to the acute angle A) [b] = AB = 24 cm

And, hypotenuse (opposite side of the right angle B) [h] = AC = 25 cm

We know that,

sin A = ^{Perpendicular(p)}/_{Hypotenuse(h)}

After substituting the values of perpendicular and hypotenuse, we get

⇒ sin A = ^{7 cm}/_{25 cm}

⇒ sin A = ⁷/₂₅

We know that, cos A = ^Base(b)/_{Hypotenuse(h)}

After substituting the values of Base(b) and Hypotenuse(h), we get

⇒ cos A = ^{24 cm}/_{25 cm}

⇒ cos A = ²⁴/₂₅

Thus, sin A = ⁷/₂₅ and cos A = ²⁴/₂₅ Answer

(ii) sin C, cos C

Solution:

Here, we have,

For acute angle C,

Hypotenuse [h] (side opposite to the right angle B) = AC = 25 cm (as calculated)

Perpendicular [p] (side opposite to the acute angle C) = AB = 24 cm (according to question)

Base [b] (side adjacent to the acute angle C) = BC = 7 cm (according to question)

Now, we know that,

sin C = ^AB/_AC

⇒ sin C= ^{Perpendicular(p)}/_{Hypotenuse(h)}

After substituting the values of perpendicular and hypotenuse, we get

⇒ sin C = ^{24 cm}/_{25 cm}

⇒ sin C = ²⁴/₂₅

Again, we know that,

cos C = ^BC/_AC

⇒ cos C= ^Base(b)/_{Hypotenuse(h)}

After substituting the values of Base(b) and Hypotenuse(h), we get

⇒ cos C = ^{7 cm}/_{25 cm}

⇒ cos C = ⁷/₂₅

Thus,

sin C = ²⁴/₂₅ and cos C = ⁷/₂₅ Answer

Question (2) In Figure, find tan P – cot R.

Solution:

(Strategy to solve the question: two sides of the given right angle triangle are given in the question. Thus after calculating the third side using Pythagoras Theorem, we can calculate the required ratio.)

Here, given,

∠ Q = 90^o i.e. right angle

Perpendicular (p) = PQ = 12 cm

Hypotenuse (h) = PR = 13 cm

Thus, Base (b) = QR =?

According to Pythagoras Theorem, we know that

(hypotenuse) ² = (perpendicular) ² + (base) ²

⇒ PR² = PQ² + QR²

⇒ (13 cm) ² = (12 cm) ² + QR²

⇒ 169 cm² = 144 cm² + QR²

⇒ QR² = 169 cm² – 144 cm²

⇒ QR² = 25 cm²

⇒ QR = (25 cm)²

⇒ QR = 5 cm = Base(b)

Now, for acute angle P,

Perpendicular (p) [the opposite of the acute angle P] = QR = 5 cm (as calculated)

Base (b) [the adjacent side of acute angle P] = PQ = 12 cm

We know that,

tan P = ^{Perpendicular(p)}/_Base(b)

⇒ = ^QR/_QP

After substituting the values of perpendicular(p) and base(b), we get

⇒ tan P = ^{5 cm}/_{12 cm}

⇒ tan P = ⁵/₁₂ - - - - (i)

For acute angle R

Perpendicular (p) [the opposite side of the acute angle R] = PQ = 12 cm

Base (b) [the adjacent side of the acute angle R] = QR = 5 cm

We know that,

cot R = ^Base(b)/_{Perpendicular(p)}

After substituting the values of base(b) and perpendicular(p), we get

⇒ cot R = ^{5 cm}/₁₂

⇒ cot R = ⁵/₁₂ - - - - (ii)

Thus,

tan P – cot R

After substituting the values of tan P and cot R from the above equations (i) & (ii), we get

= ⁵/₁₂ – ⁵/₁₂ = 0

Or, tan P – cot R = 0 Answer

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

Solution:

[Strategy to solve the question: the value of sin A is given. This means the ratio of two sides of a right-angle triangle is given. Thus, after the calculation of the third side the required trigonometric ratio can be calculated.]

Let, ABC is the right-angle triangle as given in the question.

In which ∠ C = 90^o = Right Angle

Here, given, sin A = ³/₄

We know that,

sin A = ^{Perpendicular(p)}/_{Hypotenuse(h)}

Thus,

Perpendicular (p) = BC = 3, and

Hypotenuse (h) = AB = 4

According to Pythagoras Theorem, we know that

(hypotenuse) ² = (perpendicular) ² + (base) ²

Thus for the given right angle triangle ABC

⇒ AB² = BC² + AC²

After substituting the values of AB (hypotenuse) and BC (perpendicular)

⇒ (4)² = (3) ² + AC²

⇒ AC² = 16 – 9

⇒ AC² = 7

⇒ AC = 7

⇒ Base (b) = 7

[∵ AC is the adjacent side to the acute angle A, hence it is the base for the given right angle triangle. ]

Calculation of trigonometric ratio of cosA

Now, we know that,

cos A = ^Base(b)/_{Hypotenuse(h)}

Thus, after substituting the values of base(b) and hypotenuse(h), we get

cos A = ⁷/₄

Calculation of trigonometric ratio of tanA

And we know that, tan A = ^{Perpendicular (p)}/_{Base (b)}

Thus, after substituting the values of Perpendicular(p) and base(b), we get

⇒ tan A = ³/₇

Thus, cos A = ⁷/₄ and tan A = ³/₇ Answer

Alternate Method to solve the question "If sin A = ³/₄, calculate cos A and tan A."

Given, sin A = ³/₄

we know,

cos A = 1 – sin² A

After substituting the value of sin A, we get

cos² A = 1 – (³/₄)²

⇒ cos² A = 1 – ⁹/₁₆

⇒ cos² A = ^{16 – 9}/₁₆

⇒ cos² A = ⁷/₁₆

⇒ cos A = ⁷/₁₆

⇒ cos A = ⁷/₄

we know that

tan A = ^{sin A}/_{cos A}

After substituting values of sin A and cos A, we get

tan A = ^3/₄/_⁷/4

= ³/₄ × ⁴/₇

⇒ tan A = ³/₇

Thus cos A = ⁷/₄ and tan A = ³/₇ Answer

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