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

Trigonometry is an important branch of mathematics that deals with the relation between the lengths of sides and angles of a right-angled triangle. The word trigonometry is derived from Greek words where ‘tri’ means ‘three’, ‘gon’ means ‘sides’ and ‘metron’ means ‘measure’. Hipparchus is known as the “father of trigonometry”. Trigonometry is used in various applications in our daily life. 

Right-angled triangle

Trigonometric Ratios

By using trigonometric ratios, we can find the missing angles or sides of a right-angled triangle. Sine, Cosine, tangent, cosecant, secant, and cotangent are the six trigonometric ratios or functions. Where a trigonometric ratio is depicted as the ratio between the sides of a right-angled triangle. 

The side opposite to the right angle or the longest side is known as hypotenuse. The side opposite to the given angle is known as the opposite side and the side next to the given angle is known as the adjacent side. The sine function is defined as the ratio of the opposite side to the hypotenuse. The cosine function is defined as the ratio of the adjacent side to the hypotenuse. The tangent function is defined as the ratio of the opposite side to the adjacent side. The cosecant function is reciprocal of the sine function. The secant function is reciprocal of the cosine function and the Cotangent function is reciprocal of the tangent function. Hence, sine, cosine, and tangent functions are three primary trigonometric functions.

• sin θ = Opposite side/Hypotenuse
• cos θ = Adjacent side/Hypotenuse
• tang θ = sin θ/cos θ = Opposite side/Adjacent side
• cosec θ = 1/sin θ = Hypotenuse/Opposite side
• sec θ = 1/cos θ = Hypotenuse/Adjacent side
• cot θ = 1/tan θ = Adjacent side/Opposite side = cos θ/sin θ 

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

For finding the unknown angle of the given triangle we need to use inverse trigonometric ratios. Where inverse trigonometric functions are inverse functions of the trigonometric functions. We know that, sin θ = opposite side/hypotenuse.

Now, θ = sin^-1 (opposite side/hypotenuse)

Similarly, 

• θ = cos^-1 (adjacent side/hypotenuse)
• θ = tan^-1 (opposite side/adjacent side)
• θ = cosec^-1 (hypotenuse/opposite side)
• θ = sec^-1 (hypotenuse/adjacent side)
• θ = cot ^-1 (adjacent side/opposite side)

We know that using the six trigonometric ratios we can find the missing or unknown angles and sides of a right-angled triangle. But by using the sine rule formula and cosine rule formula we can find the sides and angles of any given triangle.

Sine Rule or the Law of sines

Sine Rule or the Law of Sines is a trigonometric law that gives a relationship between the sides and angles of the triangle (non-right angled triangle). Let a, b, and c be the lengths of the three sides of a triangle ABC and A, B, and C by their respective opposite angles. Now the expression for the Sine Rule is given as,

Triangle ABC

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 or the law of cosines is used to find the unknown side or angle of a triangle. Let a, b, and c are sides, and A, B, and C are angles of the triangle ABC. Now the expressions for the Cosine rule are given as,

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

Sample Problems

Problem 1: Find the angle ∠ACB (in degrees) of right-angled triangle ABC when AB = 5 cm and AC = 13 cm.

Solution: 

Triangle ABC

Given, AB = 5 cm and AC = 13 cm.

So, we know the length of the side opposite to ∠ACB and length of the hypotenuse. Therefore, we can use the sine trigonometric function to find ∠ACB.

sin C = opposite side/ hypotenuse

⇒ sin C = AB/AC

⇒ sin C = 5/13

⇒ ∠C = sin^-1 (5/13) ⇒ ∠C = 22.61°

Hence, ∠ACB = 22.61°

Problem 2: Find ∠Z (in degrees) of right-angled triangle XYZ when XY = 6 cm and YZ = 8 cm.

Solution:

Triangle XYZ

Given, XY = 6 cm and YZ = 8 cm.

So, we know the lengths of the opposite and adjacent sides of ∠Z. Therefore, we can use tangent function to find ∠Z

We have, tan Z = opposite side/adjacent side

⇒ tan Z = XY/YZ

⇒ tan Z = 6/8 = 3/4

⇒ ∠Z = tan^-1 (3/4) ⇒ ∠Z = 36.87°

Hence, ∠Z = 36.87°

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

Solution:

Triangle ABC

Given, AB = 3 cm and AC = 5 cm

So, we the length of the adjacent side of ∠A and length of the hypotenuse. Therefore, we can use cosine function to find ∠A

cos A = adjacent side/hypotenuse

⇒ cos A = AB/AC

⇒ cos A = 3/5

⇒ ∠A = cos^-1 (3/5) ⇒ ∠A = 53.13°

Hence, ∠A = 53.13°

Problem 4: Find ∠P (in degrees) and ∠R (in degree) and the length of the hypotenuse of a right-angled triangle, if PQ = 24 cm and QR = 7 cm.

Solution:

Given data, PQ = r = 24 cm and QR = p = 7 cm

By Pythagoras theorem we have, p² + q² = r²

⇒ 24² + 7² = r²

⇒ r² = 576 + 49 = 625

⇒ r = √625 ⇒ r = 25 cm

We have, sin P = opposite side/hypotenuse

⇒ sin P = 7/25

⇒ ∠P = sin^-1 (7/25) ⇒ ∠P = 16.26°

We know that, sum of angles in a triangle = 180°

⇒ ∠P + ∠Q + ∠R = 180°

⇒ 16.26° + 90° + ∠R = 180°

⇒ ∠R = 180° – 16.26° – 90° ⇒ ∠R = 73.76°

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

Problem 5: Find ∠B (in degrees) and ∠C (in degrees), if ∠A = 30° and AB = 16 inches and BC = 12 inches.

Solution:

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

Triangle ABC

We can use the law of sines to find ∠B and ∠C.

a/sin A = b/sin B = c/sin C

Consider, a/sin A = c/sin C

⇒ 12/sin 30° = 16/sin C [sin 30° = ½]

⇒ 12/(½) = 16/sin C

⇒ sin C = 16/24 = 2/3

⇒ ∠C = sin^-1 (2/3) = 41. 81°

Sum of angles in a triangle = 180°

∠A + ∠B + ∠C = 180°

⇒ 30° + ∠B + 41.81° = 180° ⇒ ∠B = 180° – 30° – 41.81°

⇒ ∠B = 101.19°

Hence, ∠B = 101.19° and ∠C = 41.81°

Problem 6: 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 data, Area of triangle XYZ = 24 cm^2,, YZ = x = 12 cm and XY = z = 5 cm

Triangle XYZ

From the law of sines we have, Area of triangle = ½ × (x) × (z) × sin Y

⇒ 24 cm² = ½ × (12) × (5) × sin Y

⇒ sin Y = 48/60 = 4/5

⇒ ∠Y = sin^-1 (4/5)

⇒ ∠Y = 53.12°

Problem 7: Find the measure of angles (in degrees) of triangle PQR when PQ = 14 cm, QR = 6 cm, and PR = 5 cm.

Solution:

Given data, PQ = r = 5 cm, QR = p = 7 cm and PR = q = 8 cm

Triangle PQR

Here we know the lengths of the three sides of the triangle PQR. So, we can use Cosine Rule Law to find ∠Q.

We have, cos Q = (p² + r² – q²)/2pr

⇒ cos Q = {[(7)² + (5)² – (8)²]/(2 × 7 × 5)}

⇒ cos Q = {[49 + 25 – 64]/70}

⇒ cos Q = 10/70 = 1/7

⇒ ∠Q = cos^-1 (1/7) ⇒ ∠Q = 81.79°

Similarly, cos P = (q² + r² – p²)/2qr

⇒ cos P = {[(8)² + (5)² – (7)²]/(2 × 8 × 5)}

⇒ cos P = {[64 + 25 – 49]/80}

⇒ cos P = 40/80 = ½

⇒ ∠P = cos^-1 (½) ⇒ ∠P = 60°

cos R = (p² + q² – r² )/2pq

⇒ cos R = {[(7)² + (8)² – (5)²]/2 × 7 × 8}

⇒ cos R ={[49 + 64 – 25]/112}

⇒ cos R = 88/112 ⇒ cos R = 11/14

⇒ ∠R = cos^-1 (11/14) ⇒ ∠R = 38.21°

Hence, the angles of triangle PQR are ∠P = 60°, ∠Q = 81.79° & ∠R = 38.21°