Worksheet on Trigonometry Table

Angle relationships unlock the mathematical secrets of triangles beautifully. Advanced learners work with printable trigonometry table practice questions that explore sine, cosine, and tangent values for different angle measurements. These sophisticated exercises help students understand trigonometric ratios while connecting geometry to physics, engineering, and navigation applications. Students can download a free trigonometry table worksheet PDF, every value represents precise mathematical relationships.

1. A building casts a shadow of 40 m when the angle of elevation of the sun is 45^o. How tall is the building?

a) 20 m
b) 30 m
c) 40 m
d) 45 m

Answer: c)

Explanation: Here, θ = 45^o and the adjacent side (shadow length) is 40 m.
Let h be the height of the building:
Tan 45^o = $\frac{\frac{h}{}}{\frac{\mathrm{40}}{}}$
Since Tan 45^o = 1
1 = $\frac{\frac{h}{}}{\frac{\mathrm{40}}{}}$
h = 40

Hence, height of the building is 40 m.

2. Find the value of $\frac{\frac{4}{}}{\frac{3}{}}$ tan²60^o + 3cos²30^o - 2sec²30^o - 3cot²60^o.

a) $\frac{\frac{7}{}}{\frac{\mathrm{12}}{}}$
b) $\frac{\frac{\mathrm{11}}{}}{\frac{\mathrm{12}}{}}$
c) 3
d) $\frac{\frac{\mathrm{31}}{}}{\frac{\mathrm{12}}{}}$

Answer: d)

Explanation: $\frac{\frac{4}{}}{\frac{3}{}}$ tan²60^o + 3cos²30^o - 2sec²30^o - 3cot²60^o
= $\frac{\frac{4}{}}{\frac{3}{}}$ x √3² + 3 ($\frac{\sqrt{3}{}^{}}{2}$)² - 2 ($\frac{2{}^{}}{\sqrt{3}}$)² - 3 ($\frac{1{}^{}}{\sqrt{3}}$)² 
= $\frac{\frac{4}{}}{\frac{3}{}}$ x 3 + 3 x $\frac{\frac{3}{}}{\frac{4}{}}$  - 2 x $\frac{\frac{4}{}}{\frac{3}{}}$ - 3 x $\frac{\frac{1}{}}{\frac{3}{}}$
= 4 + $\frac{\frac{9}{}}{\frac{4}{}}$ - $\frac{\frac{8}{}}{\frac{3}{}}$ - 1
= $\frac{\frac{\mathrm{48}}{}}{\frac{\mathrm{12}}{}}$ + $\frac{\frac{\mathrm{27}}{}}{\frac{\mathrm{12}}{}}$ - $\frac{\frac{\mathrm{32}}{}}{\frac{\mathrm{12}}{}}$ - $\frac{\frac{\mathrm{12}}{}}{\frac{\mathrm{12}}{}}$
= $\frac{\frac{\mathrm{48\; +\; 27\; -\; 32\; -\; 12}}{}}{\frac{\mathrm{12}}{}}$
= $\frac{\frac{\mathrm{31}}{}}{\frac{\mathrm{12}}{}}$

3. If sin (x + y) = 1 and cos (x - y) = $\frac{\sqrt{3}{}^{}}{2}$, find x and y.

a) x = 60^o and y = 45^o
b) x = 30^o and y = 60^o
c) x = 60^o and y = 30^o
d) x = 45^o and y = 30^o

Answer: c)

Explanation: sin (x + y) = 1
⇒sin (x + y) = sin 90^o, [since sin 90^o = 1]
⇒x + y = 90^o ----- (1)

cos (x - y) = $\frac{\sqrt{3}{}^{}}{2}$
⇒ cos (x - y) = cos 30^o
⇒ x - y = 30^o ----- (2)

By solving equations (1) and (2), we will get
x = 60^o and y = 30^o

4. Evaluate: sin 60^ocos 30^o + sin 30^ocos 60^o

a) 1
b) $\frac{\frac{1}{}}{\frac{2}{}}$
c) $\frac{\sqrt{3}{}^{}}{2}$
d) √3

Answer: a)

Explanation: We know that sin 60^o = $\frac{\sqrt{3}{}^{}}{2}$ , cos 30^o = $\frac{\sqrt{3}{}^{}}{2}$, sin 30^o = $\frac{\frac{1}{}}{\frac{2}{}}$, cos 60^o = $\frac{\frac{1}{}}{\frac{2}{}}$

Substituting the values in the given expression: = $\frac{\sqrt{3}{}^{}}{2}$ . $\frac{\sqrt{3}{}^{}}{2}$ + $\frac{\frac{1}{}}{\frac{2}{}}$ . $\frac{\frac{1}{}}{\frac{2}{}}$
                                                                      = $\frac{\frac{3}{}}{\frac{4}{}}$ + $\frac{\frac{1}{}}{\frac{4}{}}$
                                                                      = 1

5. If tan 2A = cot (A-18^o), where 2A is an acute angle, find the value of A.

a) 30^o
b) 72^o
c) 36^o
d) 60^o

Answer: c)

Explanation: We know that cot x = tan (90^o- x)
Tan 2A = Tan (90^o− (A − 18^o))
Tan 2A = Tan (108^o − A)
      2A = 108^o − A
        A = $\frac{\frac{{\mathrm{108}}^{\mathrm{o}}}{}}{\frac{3}{}}$
           = 36^o

6. Evaluate:$\frac{\frac{{\mathrm{1-Tan}}^2}{{\mathrm{45}}^{\mathrm{o}}}}{\frac{{\mathrm{1+Tan}}^2}{{\mathrm{45}}^{\mathrm{o}}}}$

a) 0
b) 1
c) √3
d) -1

Answer: a)

Explanation: We know that Tan 45^o = 1
Therefore, $\frac{\frac{{\mathrm{1-Tan}}^2}{{\mathrm{45}}^{\mathrm{o}}}}{\frac{{\mathrm{1+Tan}}^2}{{\mathrm{45}}^{\mathrm{o}}}}$ = $\frac{\frac{\mathrm{1-1}}{}}{\frac{\mathrm{1+1}}{}}$
               = $\frac{\frac{0}{}}{\frac{2}{}}$
               = 0

7. If sec 4A = cosec ( A – 20^o), where 4A is an acute angle, find the value of A.

a) 28^o
b) 20^o
c) 44^o
d) 22^o

Answer: d)

Explanation: We know that sec x = cosec (90^o- x)
Sec 4A = cosec (A - 20^o)
Cosec (90^o- 4A) = cosec (A - 20^o)
           90^o - 4A  = A - 20^o
                 110^o  = 5A
                     A  = 22^o

8. Find the value of tan 10^o tan 15^o tan 75^o tan 80^o

a) 0
b) 1
c) √3
d) $\frac{1{}^{}}{\sqrt{3}}$

Answer: b)

Explanation: We know that tan (90^o- θ) = cot θ and cot θ tan θ = 1
= tan 10^o tan 15^o tan 75^o tan 80^o
= tan(90 - 80)^o tan 15^o tan(90 - 15)^o tan 80^o
= cot 80^o tan 15^o cot 15^o tan 80^o
=(cot 80^o × tan 80^o) ( cot 15^o × tan 15^o)
= 1

9. A person is standing 20 m away from the base of a building and observes the top of the building at an angle of elevation of 60^o. What is the height of the building?

a) 17.34 m
b) 34.64 m
c) 20 m
d) 30 m

Answer: a)

Explanation: Base = 20 m
Angle of elevation = 60^o
Let h be the height of the building.
As tan θ = $\frac{\frac{\mathrm{Perpendicular}}{}}{\frac{\mathrm{Base}}{}}$
⇒ tan 60^o = $\frac{\frac{h}{}}{\frac{\mathrm{20}}{}}$
⇒ h = 20 x tan 60^o
We know that tan 60^o = √3
h = 20 x √3 ≈ 34.64 m

Therefore, the height of the building is approximately 34.64 m.

10. If a cos 45^o = b sec 60^o, what is the value of $\frac{\frac{a^4}{}}{\frac{b^4}{}}$?

a) 64
b) 32
c) 216
d) 81

Answer: a)

Explanation: We know that cos 45^o = $\frac{1{}^{}}{\sqrt{2}}$ and sec 60^o = 2
Given: a cos 45^o = b sec 60^o
So, $\frac{\frac{a}{}}{\frac{b}{}}$ = $\frac{\frac{{\mathrm{Sec\; 60}}^{\mathrm{o}}}{}}{\frac{{\mathrm{Cos\; 45}}^{\mathrm{o}}}{}}$
$\frac{\frac{a}{}}{\frac{b}{}}$ = $\frac{\frac{2}{}}{\frac{\mathrm{1\; /\; \surd 2}}{}}$
$\frac{\frac{a}{}}{\frac{b}{}}$ = $\frac{\frac{\mathrm{2\surd 2}}{}}{\frac{1}{}}$

Now, $\frac{\frac{a^4}{}}{\frac{b^4}{}}$ = ($\frac{\frac{{\mathrm{2\surd 2}}^{}}{}}{\frac{1}{}}$) ⁴ 
$\frac{\frac{a^4}{}}{\frac{b^4}{}}$ = $\frac{\frac{\mathrm{64}}{}}{\frac{1}{}}$ = 64