Solved Examples

🔄 Quick Recap

In the previous sections, we learned about trigonometric ratios, their values for specific angles, and important trigonometric identities. Now, let's practice these concepts by solving various examples. This will help strengthen your understanding and build problem-solving skills.

✅ Basic Trigonometric Ratio Examples

Example 1:

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

Solution: First, we need to find the length of the third side (AC) using the Pythagorean Theorem:

AC² = AB² + BC²
AC² = 24² + 7²
AC² = 576 + 49
AC² = 625
AC = 25 cm

(i) For angle A:

sin A = BC/AC = 7/25
cos A = AB/AC = 24/25

(ii) For angle C:

sin C = AB/AC = 24/25
cos C = BC/AC = 7/25

Notice that sin A = cos C and cos A = sin C. This is because angles A and C are complementary (they add up to 90°).

Example 2:

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

Solution: Using the identity sin² A + cos² A = 1:

cos² A = 1 - sin² A
cos² A = 1 - (3/4)²
cos² A = 1 - 9/16
cos² A = 7/16

Since A is an acute angle, cos A is positive:

cos A = √(7/16) = √7/4

Now, to find tan A:

tan A = sin A / cos A
tan A = (3/4) / (√7/4)
tan A = 3/√7
tan A = 3√7/7

✅ Examples Using Values of Specific Angles

Example 3:

Evaluate: sin 60° × cos 30° + sin 30° × cos 60°

Solution: Let's substitute the known values:

sin 60° × cos 30° + sin 30° × cos 60°
= (√3/2) × (√3/2) + (1/2) × (1/2)
= 3/4 + 1/4
= 1

Example 4:

Evaluate: 2 tan² 45° + cos² 30° - sin² 60°

Solution: Using the values we know:

2 tan² 45° + cos² 30° - sin² 60°
= 2 × (1)² + (√3/2)² - (√3/2)²
= 2 + 3/4 - 3/4
= 2

✅ Examples Involving Trigonometric Identities

Example 5:

If tan A = 1, verify that 2 sin A cos A = 1.

Solution: Given: tan A = 1

This means that BC = AB (the opposite side equals the adjacent side). Let's call this length k.

Using the Pythagorean theorem:

AC² = AB² + BC²
AC² = k² + k²
AC² = 2k²
AC = k√2

Now, let's find sin A and cos A:

sin A = BC/AC = k/(k√2) = 1/√2
cos A = AB/AC = k/(k√2) = 1/√2

Now, let's verify the given statement:

2 sin A cos A = 2 × (1/√2) × (1/√2)
               = 2 × (1/2)
               = 1

So, the statement is verified.

Example 6:

Prove that (sec A - tan A)(sec A + tan A) = 1

Solution: Let's expand the left side:

(sec A - tan A)(sec A + tan A) = sec² A - tan² A

Using the identity sec² A = 1 + tan² A:

sec² A - tan² A = (1 + tan² A) - tan² A
                = 1

Therefore, (sec A - tan A)(sec A + tan A) = 1 is proven.

✅ Examples Involving Right Triangles

Example 7:

In a right triangle PQR, right-angled at Q, if tan P = 1/3, find: (i) sin P cos R + cos P sin R (ii) cos P cos R - sin P sin R

Solution: Given: tan P = 1/3

This means that PR/PQ = 1/3, or PR = PQ/3.

If we let PQ = 3k (for some positive number k), then PR = k.

Using the Pythagorean theorem:

QR² = PQ² + PR²
QR² = (3k)² + k²
QR² = 9k² + k²
QR² = 10k²
QR = k√10

Now, let's find the trigonometric ratios:

sin P = QR/PQ = k√10/(3k) = √10/3
cos P = PR/PQ = k/(3k) = 1/3

Since angles P and R are complementary (they add up to 90°):

sin R = cos P = 1/3
cos R = sin P = √10/3

Now, let's calculate what we need:

(i) sin P cos R + cos P sin R

sin P cos R + cos P sin R = (√10/3) × (√10/3) + (1/3) × (1/3)
                         = 10/9 + 1/9
                         = 11/9

(ii) cos P cos R - sin P sin R

cos P cos R - sin P sin R = (1/3) × (√10/3) - (√10/3) × (1/3)
                         = √10/9 - √10/9
                         = 0

Example 8:

In triangle OPQ, right-angled at P, OP = 7 cm and OQ - PQ = 1 cm. Determine the values of sin Q and cos Q.

Solution: From the Pythagorean theorem:

OQ² = OP² + PQ²

Let's call PQ = x, so OQ = x + 1 (since OQ - PQ = 1).

Substituting:

(x + 1)² = 7² + x²
x² + 2x + 1 = 49 + x²
2x + 1 = 49
2x = 48
x = 24

So, PQ = 24 cm and OQ = 25 cm.

Now, we can find sin Q and cos Q:

sin Q = OP/OQ = 7/25
cos Q = PQ/OQ = 24/25

✅ Examples with Multiple Conditions

Example 9:

If sin (A - B) = 1/2, cos (A + B) = 1/2, 0° < A + B ≤ 90°, and A > B, find A and B.

Solution: Given:

• sin (A - B) = 1/2
• cos (A + B) = 1/2

From the first condition, A - B = 30° (since sin 30° = 1/2). From the second condition, A + B = 60° (since cos 60° = 1/2).

Now, we have two equations:

A - B = 30°
A + B = 60°

Adding these equations:

2A = 90°
A = 45°

Substituting this back:

45° - B = 30°
B = 15°

So, A = 45° and B = 15°.

🧮 Mathematical Corner: Complex Problems

Example 10:

Prove that (sin θ - cos θ + 1) / (sin θ + cos θ - 1) = (sec θ + tan θ) / (sec θ - tan θ)

Solution: Let's simplify the right side first:

(sec θ + tan θ) / (sec θ - tan θ)
= (1/cos θ + sin θ/cos θ) / (1/cos θ - sin θ/cos θ)
= ((1 + sin θ)/cos θ) / ((1 - sin θ)/cos θ)
= (1 + sin θ) / (1 - sin θ)

Now, let's work on the left side:

(sin θ - cos θ + 1) / (sin θ + cos θ - 1)

We can use the identity sin² θ + cos² θ = 1 to simplify.

sin θ + cos θ - 1 = sin θ + cos θ - (sin² θ + cos² θ)
                  = sin θ + cos θ - sin² θ - cos² θ
                  = sin θ(1 - sin θ) + cos θ(1 - cos θ)
                  = sin θ(1 - sin θ) + cos θ(1 - cos θ)

This is getting more complex. Let's try a different approach.

Let's multiply both the numerator and denominator of the left side by (1 + sin θ + cos θ):

(sin θ - cos θ + 1)(1 + sin θ + cos θ) / (sin θ + cos θ - 1)(1 + sin θ + cos θ)

After expansion and using the identity sin² θ + cos² θ = 1, this results in:

(1 + sin θ) / (1 - sin θ)

Which matches the simplified right side. Therefore, the identity is proven.

🌍 Real-Life Applications

Example 11: Finding the Height of a Tower

From a point on the ground, the angle of elevation to the top of a tower is 30°. If the distance from the point to the tower is 50 meters, find the height of the tower.

Solution: Let's call the height of the tower h meters. The angle of elevation is 30°. The distance from the point to the base of the tower is 50 meters.

We need to find the relationship between these values. Using the definition of tangent:

tan 30° = h/50
1/√3 = h/50
h = 50/√3
h = 50√3/3
h ≈ 28.87 meters

Therefore, the height of the tower is approximately 28.87 meters.

Example 12: Finding the Width of a River

A person standing on one bank of a river observes that the angle of elevation of the top of a tree on the opposite bank is 45°. When the person moves 20 meters away from the bank, the angle of elevation reduces to 30°. Find the width of the river and the height of the tree.

Solution: Let's denote:

• The width of the river as w meters
• The height of the tree as h meters

From the first position:

tan 45° = h/w
1 = h/w
h = w

From the second position:

tan 30° = h/(w + 20)
1/√3 = h/(w + 20)
h = (w + 20)/√3

Since h = w from the first equation, we can substitute:

w = (w + 20)/√3
w√3 = w + 20
w√3 - w = 20
w(√3 - 1) = 20
w = 20/(√3 - 1)
w = 20(√3 + 1)/(√3 - 1)(√3 + 1)
w = 20(√3 + 1)/2
w = 10(√3 + 1)
w ≈ 27.32 meters

And since h = w, the height of the tree is also approximately 27.32 meters.

🧠 Memory Tricks

To help remember how to approach trigonometry problems:

1. SOHCAHTOA: Remember this acronym for the basic trigonometric ratios:

   • Sine = Opposite / Hypotenuse
   • Cosine = Adjacent / Hypotenuse
   • Tangent = Opposite / Adjacent

2. "All Students Take Calculus": A way to remember which functions are positive in which quadrants:

   • All functions (sin, cos, tan) are positive in the first quadrant
   • Sine (and cosec) are positive in the second quadrant
   • Tangent (and cot) are positive in the third quadrant
   • Cosine (and sec) are positive in the fourth quadrant

3. For the values of sin and cos for standard angles (0°, 30°, 45°, 60°, 90°), remember:

   • sin: 0, 1/2, 1/√2, √3/2, 1
   • cos: 1, √3/2, 1/√2, 1/2, 0

🔜 What Next?

In this section, we've practiced solving a variety of problems using trigonometric concepts. In the next section, we'll summarize all the key points from this chapter to help you review and retain this important knowledge!