Trigonometry for RRB NTPC: values, identities and heights

Trigonometry questions in RRB NTPC are mostly quick. If you know the standard values and three identities, most of them are direct substitution. The rest are heights-and-distances questions, which are really right-triangle questions from geometry with an angle given instead of a side.

The whole chapter rests on one right triangle. Keep that picture in mind and the formulas stop being a list to memorise.

The six ratios

In a right triangle, for an angle θ (not the right angle):

Also, tan θ = sin θ ÷ cos θ, because the hypotenuse cancels.

Standard values

Why these values: a 45° right triangle has two equal legs, so with legs 1 and 1 the hypotenuse is √2 and sin 45° = 1/√2. Cutting an equilateral triangle of side 2 in half gives a 30-60-90 triangle with sides 1, √3 and 2, which gives every 30° and 60° value.

The three identities

• sin²θ + cos²θ = 1
• sec²θ − tan²θ = 1
• cosec²θ − cot²θ = 1

Why: by Pythagoras, opposite² + adjacent² = hypotenuse². Divide every term by hypotenuse² and you get the first identity. Divide by adjacent² instead and you get the second; by opposite², the third.

A useful consequence: since sec²θ − tan²θ = (sec θ + tan θ)(sec θ − tan θ) = 1, the two brackets are reciprocals. If one is given, the other is 1 divided by it. The same holds for cosec θ + cot θ and cosec θ − cot θ.

Complementary angles

• sin (90° − θ) = cos θ, and cos (90° − θ) = sin θ
• tan (90° − θ) = cot θ
• sec (90° − θ) = cosec θ

Why: in a right triangle the two acute angles add to 90°. The side opposite one angle is adjacent to the other, so the sin of one is the cos of the other.

So sin 30° = cos 60°, and tan 20° × tan 70° = tan 20° × cot 20° = 1. When a question pairs angles that add to 90°, look for this.

Finding other ratios from one

If one ratio is given as a fraction, draw the triangle, find the third side by Pythagoras (often a triplet) and read off anything else. For sin θ = 3/5, the sides are 3 and 5, the third is 4.

Heights and distances

• Angle of elevation: looking up from the horizontal to an object above.
• Angle of depression: looking down from the horizontal to an object below. It equals the angle of elevation from the lower point, because the two horizontals are parallel.

When a question gives a height and a horizontal distance, use tan. It relates exactly those two sides.

Worked questions at NTPC level

Q1. Find sin²30° + cos²60° + tan²45°.

1/4 + 1/4 + 1 = 3/2.

Q2. If sin θ = 3/5, find cos θ and tan θ.

Opposite 3, hypotenuse 5, so adjacent 4.
cos θ = 4/5; tan θ = 3/4.

Q3. Find sin 35° ÷ cos 55° + tan 10° × tan 80°.

cos 55° = sin 35°, so the first term is 1. tan 80° = cot 10°, so the second is 1.
Total = 2.

Q4. If sec θ + tan θ = 2, find sec θ − tan θ and sec θ.

sec θ − tan θ = 1 ÷ 2 = 1/2.
Adding the two equations: 2 sec θ = 5/2, so sec θ = 5/4.

Q5. A pole 6 m tall casts a shadow 2√3 m long. Find the angle of elevation of the sun.

tan θ = 6 ÷ 2√3 = √3, so θ = 60°.

Q6. From a point on the ground, the angle of elevation of a tower's top is 30°. After walking 20 m towards the tower, it becomes 60°. Find the height.

Let the height be h and the final distance x. Then h = x√3, and h = (x + 20)/√3.
So 3x = x + 20, giving x = 10. Height = 10√3 m, about 17.3 m.

Common mistakes

• Swapping the 30° and 60° values of sin and cos.
• Using sin when a horizontal distance is given. Height with base means tan.
• Treating tan 90° as a number. It is not defined.
• Forgetting that complementary pairs must add to exactly 90°. sin 40° and cos 40° are not equal.
• Squaring the angle. sin²θ means (sin θ)², not sin (θ²).

Practice set

1. Find sin 60° × cos 30°.
2. Find tan 60° ÷ tan 30°.
3. If cos θ = 5/13, find sin θ.
4. At a point 25 m from the foot of a building, the angle of elevation of the top is 45°. Find the height.
5. Find sec²θ − tan²θ.
6. If cosec θ − cot θ = 1/3, find cosec θ + cot θ.
7. A ladder makes an angle of 60° with the ground, and its foot is 4 m from the wall. Find the length of the ladder.
8. Simplify (1 − sin²θ) × sec²θ.

Answers:

1. 3/4. √3/2 × √3/2.
2. 3. √3 ÷ (1/√3).
3. 12/13. 5-12-13 triplet.
4. 25 m. tan 45° = 1.
5. 1. Identity.
6. 3. The two expressions are reciprocals.
7. 8 m. cos 60° = 4 ÷ L, so L = 4 ÷ 1/2.
8. 1. 1 − sin²θ = cos²θ, and cos²θ × sec²θ = 1.

What to do next

• Write the values table from memory every day this week, then build it from the two triangles to check.
• Solve 15 identity and complementary-angle questions, spotting the 90° pairs first.
• Draw a figure for 10 heights-and-distances questions, marking the right angle and the given angle.
• Move on to data interpretation to finish the Maths syllabus.