How to find an angle in a right-angled triangle?

A right-angle triangle is a type of triangle that has one angle measuring exactly 90 degrees (90°).

To find an angle in a right-angled triangle, trigonometric ratios are used. These relate the angles to the sides of the triangle.

Trigonometric Ratios

Trigonometric ratios are used to find unknown sides or angles in a right-angled triangle. The six ratios are sine, cosine, tangent, cosecant, secant, and cotangent.

In a Right-Angled Triangle:

• Hypotenuse: The longest side, opposite the right angle
• Opposite Side: The side opposite the given angle (θ)
• Adjacent Side: The side next to the given angle (θ)

Primary Trigonometric Ratios:

• sin θ = Opposite / Hypotenuse
• cos θ = Adjacent / Hypotenuse
• tan θ = Opposite / Adjacent

Reciprocal Trigonometric Ratios:

• cosec θ = Hypotenuse / Opposite
• sec θ = Hypotenuse / Adjacent
• cot θ = Adjacent / Opposite

Finding Angles in a Right-Angled Triangle

To find an unknown angle in a right-angled triangle, we use inverse trigonometric functions. These functions help determine the angle when the sides of the triangle are known.

• θ = sin⁻¹ (Opposite / Hypotenuse)
• θ = cos⁻¹ (Adjacent / Hypotenuse)
• θ = tan⁻¹ (Opposite / Adjacent)
• θ = cosec⁻¹ (Hypotenuse / Opposite)
• θ = sec⁻¹ (Hypotenuse / Adjacent)
• θ = cot⁻¹ (Adjacent / Opposite)

Understanding sin θ and sin⁻¹ θ

• sin θ (sine) takes an angle and gives the ratio: Opposite / Hypotenuse
• sin⁻¹ (inverse sine) gives the angle θ from the ratio (Opposite / Hypotenuse).

Note: Trigonometric ratios are used for right-angled triangles. For other triangles, the Sine Rule and Cosine Rule are used.

Sine Rule (Law of Sines)

The Sine Rule shows the relationship between the sides and angles of any triangle (not necessarily right-angled).
In a triangle ABC, where sides a, b, c are opposite angles A, B, C respectively:

sin A/a = sin B/b = sin C/c (or) a/sin A = b/sin B = c/sin C

Cosine Rule or the law of cosines

The Cosine Rule is used to find an unknown side or angle in any triangle. In a triangle ABC, where sides a, b, c are opposite angles A, B, C respectively:

a² = b² + c² - 2ab cos A (or) cos A = (b² + c² - a²)/2bc

b² = a² + c² - 2ac cos B (or) cos B = (a² + c² - b²)/2ac

c² = a² + b² - 2ab cos C (or) cos C = (a² + b² - c²)/2ab

Solved Examples

Problem 1: Find ∠Z (in degrees) of right-angled triangle XYZ when XY = 6 cm and YZ = 8 cm.

Solution:

Given XY = 6 cm and YZ = 8 cm.
Since we know the opposite and adjacent sides, we use the tangent function:

tan Z = XY / YZ = 6 / 8 = 3 / 4
⇒ ∠Z = tan⁻¹ (3 / 4) = 36.87°

Hence, ∠Z = 36.87°.

Problem 2: Find ∠A (in degrees) of right-angled triangle ABC when AB = 3 cm and AC = 5 cm.

Solution:

Given AB = 3 cm and AC = 5 cm.
Since we know the adjacent side and hypotenuse, we use the cosine function:

cos A = AB / AC = 3 / 5
⇒ ∠A = cos⁻¹ (3 / 5) = 53.13°

Hence, ∠A = 53.13°.

Problem 3: Find ∠P (in degrees), ∠R (in degrees), and the length of the hypotenuse of a right-angled triangle PQR, right-angled at Q, if PQ = 24 cm and QR = 7 cm.

Solution:

Given PQ = 24 cm and QR = 7 cm.

By Pythagoras theorem:
PR² = PQ² + QR² = 24² + 7² = 576 + 49 = 625
⇒ PR = 25 cm

Using sine:
sin P = QR / PR = 7 / 25
⇒ ∠P = sin⁻¹ (7 / 25) = 16.26°

Using angle sum property:
∠R = 180° − 90° − 16.26° = 73.76°

Hence, ∠P = 16.26°, ∠R = 73.76°, and PR = 25 cm.

Problem 4: Find ∠B (in degrees) and ∠C (in degrees), if ∠A = 30° and AB = 16 inches and BC = 12 inches.

Solution:

Given ∠A = 30°, AB = 16 inches, and BC = 12 inches.

Using the Sine Rule:
a / sin A = c / sin C

⇒ 12 / sin 30° = 16 / sin C
⇒ 12 / (1/2) = 16 / sin C
⇒ 24 = 16 / sin C
⇒ sin C = 16 / 24 = 2 / 3

⇒ ∠C = sin⁻¹ (2 / 3) = 41.81°

Using angle sum property:
∠B = 180° − 30° − 41.81° = 108.19°

Hence, ∠C = 41.81° and ∠B = 108.19°.

Problem 5: Find the measure of ∠Y (in degrees), if the area of triangle XYZ = 24 cm² and YZ = 12 cm and XY = 5 cm.

Solution:

Given area = 24 cm², YZ = 12 cm, and XY = 5 cm.

Using area formula:
Area = ½ × (YZ) × (XY) × sin Y

⇒ 24 = ½ × 12 × 5 × sin Y
⇒ 24 = 30 sin Y
⇒ sin Y = 24 / 30 = 4 / 5

⇒ ∠Y = sin⁻¹ (4 / 5) = 53.13°

Problem 6: Find the measures of the angles (in degrees) of triangle PQR if PQ = 5 cm, QR = 7 cm, and PR = 8 cm.

Solution:

Given PQ = 5 cm, QR = 7 cm, and PR = 8 cm.

Using Cosine Rule:

cos Q = (p² + r² − q²) / 2pr
= (7² + 5² − 8²) / (2 × 7 × 5)
= (49 + 25 − 64) / 70 = 10 / 70 = 1 / 7

⇒ ∠Q = cos⁻¹ (1 / 7) = 81.79°

Similarly,

cos P = (q² + r² − p²) / 2qr
= (8² + 5² − 7²) / (2 × 8 × 5)
= (64 + 25 − 49) / 80 = 40 / 80 = 1 / 2

⇒ ∠P = cos⁻¹ (1 / 2) = 60°

cos R = (p² + q² − r²) / 2pq
= (7² + 8² − 5²) / (2 × 7 × 8)
= (49 + 64 − 25) / 112 = 88 / 112 = 11 / 14

⇒ ∠R = cos⁻¹ (11 / 14) = 38.21°