Draw the right-angled triangle and determine the third side.

sin A = $\frac{3}{5}$

sin(θ) = $\frac{\mathrm{Opposite}}{\mathrm{Hypothenuse}}$

In sin A = $\frac{3}{5}$, the opposite side of A is 3 and hypothenuse side is 5.

From pythagoras theorem

The longest side (side opposite/facing the right-angle) is the hypothenuse (c)

From the triangle, the side facing the right-angle is y, hence y is the hypothenuse (c)

b² + 3² = 5²

b² + 9 = 25

b² = 25 - 9

b² = 16

Take square root of both sides.

b = $\sqrt{\mathrm{16}}$ = 4

TOA

tan(θ) = $\frac{\mathrm{Opposite}}{\mathrm{Adjacent}}$

Side opposite to A is 3 and adjacent to A is 4

tan A = $\frac{3}{4}$

CAH

cos(θ) = $\frac{\mathrm{Adjacent}}{\mathrm{Hypothenuse}}$

Hypothenuse side of A is 5

cos A = $\frac{4}{5}$

tan A - cos A = $\frac{3}{4}$ - $\frac{4}{5}$

tan A - cos A = $\frac{\mathrm{5\; x\; 3\; -\; 4\; x\; 4}}{20}$

tan A - cos A = $\frac{\mathrm{15\; -\; 16}}{20}$

tan A - cos A = -$\frac{1}{20}$

tan(θ) = $\frac{\mathrm{Opposite}}{\mathrm{Adjacent}}$

tan x = $\frac{12}{5}$ ⇒ the opposite side to x is 12 and the adjacent side to x is 5

Draw your right-angle triangle and get the third side and use the SOHCAHTOA to find the sin x and cos x and multiply them.

From Pythagorean theorem,

The hypothenuse (c) is the longest side. The side facing the right-angle (∟)

c² = 5² + 12²

c² = 25 + 144

c² = 169

Take the square root of both sides.

c = $\sqrt{\mathrm{169}}$ = 13

∴ the hypotenuse = 13

sin(θ) = $\frac{\mathrm{Opposite}}{\mathrm{Hypothenuse}}$

sin x = $\frac{5}{13}$

cos(θ) = $\frac{\mathrm{Adjacent}}{\mathrm{Hypothenuse}}$

cos x = $\frac{12}{13}$

sin x cos x = $\frac{5}{13}$ x $\frac{12}{13}$

sin x cos x = $\frac{\mathrm{5\; x\; 12}}{\mathrm{13\; x\; 13}}$

sin x cos x = $\frac{60}{169}$

Let the width of the gutter = w

tan(θ) = $\frac{\mathrm{Opposite}}{\mathrm{Adjacent}}$

tan 35° = $\frac{15}{\mathrm{w}}$

Multiply both sides by w

w x tan 35° = 15

Divide both sides by tan 35°

w = $\frac{15}{\mathrm{tan\; 35°}}$ = 21.42 cm

Tan 30^o = $\frac{1}{\sqrt{3}}$

sin 60^o = $\frac{\sqrt{3}}{2}$

tan 30^o - 2sin 60^o = $\frac{1}{\sqrt{3}}$ - 2 x $\frac{\sqrt{3}}{2}$

Note: The 2 dividing $\sqrt{3}$ cancels the 2 multiply.

tan 30^o - 2sin 60^o = $\frac{1}{\sqrt{3}}$ - $\sqrt{3}$

Every number is being divided by 1.

$\sqrt{3}$ = $\frac{\sqrt{3}}{1}$

tan 30^o - 2sin 60^o = $\frac{1}{\sqrt{3}}$ - $\frac{\sqrt{3}}{1}$

Simplify the fraction.

tan 30^o - 2sin 60^o = $\frac{\mathrm{1\; -}\sqrt{3}x\sqrt{3}}{\sqrt{3}}$

From indices, $\sqrt{a}$ x $\sqrt{a}$ = a

Hence $\sqrt{3}$ x $\sqrt{3}$ = 3

tan 30^o - 2sin 60^o = $\frac{\mathrm{1\; -}3}{\sqrt{3}}$

tan 30^o - 2sin 60^o = $\frac{-2}{\sqrt{3}}$

Multiply both the numerator and the denominator by $\sqrt{3}$

tan 30^o - 2sin 60^o = $\frac{-2x\sqrt{3}}{\sqrt{3}x\sqrt{3}}$

$\sqrt{3}$ x $\sqrt{3}$ = 3

tan 30^o - 2sin 60^o = $\frac{-2\sqrt{3}}{3}$

tan θ = $\frac{3}{4}$

The opposite side to θ is 3.

The adjacent side to θ is 4.

Draw the right angle and find the other side.

From pythagoras theorem

c² = 3² + 4²

c² = 9 + 16

c² = 25

Take the square root of both sides.

c = $\sqrt{\mathrm{25}}$ = 5

cos(θ) = $\frac{\mathrm{Adjacent}}{\mathrm{Hypothenuse}}$

The adjacent side to θ is 4.

The hypotenuse side to θ is 5.

cos θ = $\frac{4}{5}$

Use the sign of θ in the various quadrants.

cos θ in the range 180° < θ < 270° is negative.

cos θ = - $\frac{4}{5}$

sin x = $\frac{3}{5}$

SOH

sin(θ) = $\frac{\mathrm{Opposite}}{\mathrm{Hypothenuse}}$

The opposite side to x = 3 and the hypotenuse = 5

Draw the right angle and assign the values.

From pythagoras theorem

3² + b² = 5²

b² = 5² - 3²

b² = 25 - 9

b² = 16

Take the square root of both sides.

b = $\sqrt{\mathrm{16}}$ = 4

TOA

tan(θ) = $\frac{\mathrm{Opposite}}{\mathrm{Adjacent}}$

The opposite side to x is 3

The adjacent side to x is 4

tan x = $\frac{3}{4}$

CAH

cos(θ) = $\frac{\mathrm{Adjacent}}{\mathrm{Hypothenuse}}$

The adjacent side to x is 4

The hypotenuse side to x is 5

cos x = $\frac{4}{5}$

tan x + 2 cos x = $\frac{3}{4}$ + 2 x $\frac{4}{5}$

tan x + 2 cos x = $\frac{3}{4}$ + $\frac{8}{5}$

tan x + 2 cos x = $\frac{\mathrm{5\; x\; 3\; +\; 4\; x\; 8}}{20}$

tan x + 2 cos x = $\frac{\mathrm{15\; +\; 32}}{20}$

tan x + 2 cos x = $\frac{47}{20}$ = 2$\frac{7}{20}$

tan(θ) = $\frac{\mathrm{Opposite}}{\mathrm{Adjacent}}$

Opposite side of 52° is 32 cm

Adjacent side of 52° is |XZ|

tan 52° = $\frac{32}{\mathrm{|XZ|}}$

|XZ| X tan 52° = 32

Divide both sides by tan 52°

|XZ| = $\frac{32}{\mathrm{tan\; 52°}}$ = 25 cm

tan(θ) = $\frac{\mathrm{Opposite}}{\mathrm{Adjacent}}$

tan x = $\frac{3}{4}$ means the opposite side of x is 3 and the adjacent side is 4

Use this information to draw a right-angle triangle and calculate th hypotenuse.

From pythagoras theorem

The longest side (side opposite/facing the right-angle) is the hypotenuse (c)

c² = 3² + 4²

c² = 9 + 16

c² = 25

Take square root of both sides.

c = $\sqrt{\mathrm{25}}$ = 5

SOH

sin(θ) = $\frac{\mathrm{Opposite}}{\mathrm{Hypothenuse}}$

Opposite side of x = 3

Hypothenuse = 5

sin x = $\frac{3}{5}$

CAH

cos(θ) = $\frac{\mathrm{Adjacent}}{\mathrm{Hypothenuse}}$

Adjacent side of x = 4

cos x = $\frac{4}{5}$

$\frac{\mathrm{cos\; x}}{\mathrm{2sin\; x}}$ = $\frac{4}{5}$ ÷ (2 x $\frac{3}{5}$)

$\frac{\mathrm{cos\; x}}{\mathrm{2sin\; x}}$ = $\frac{4}{5}$ ÷ $\frac{\mathrm{2\; x\; 3}}{5}$

$\frac{\mathrm{cos\; x}}{\mathrm{2sin\; x}}$ = $\frac{4}{5}$ ÷ $\frac{6}{5}$

Reciprocate the fraction at the right of the ÷ and change ÷ to multiplication (x)

Note: reciprocate means the numerator becomes the denominator and the denominator becomes the numerator.

$\frac{\mathrm{cos\; x}}{\mathrm{2sin\; x}}$ = $\frac{4}{5}$ x $\frac{5}{6}$ = $\frac{4}{6}$ = $\frac{2}{3}$

$\frac{\mathrm{cos\; 65°}}{\mathrm{sin\; 25°}}$ + $\frac{\mathrm{sin\; 35°}}{\mathrm{cos\; 55°}}$ = $\frac{\mathrm{cos\; 65°}}{\mathrm{sin\; 25°}}$ + $\frac{\mathrm{sin\; 35°}}{\mathrm{cos\; 55°}}$

cos 65° = cos (90° - 25°) = sin 25°

sin 35° = sin (90° - 55°) = cos 55°

$\frac{\mathrm{cos\; 65°}}{\mathrm{sin\; 25°}}$ + $\frac{\mathrm{sin\; 35°}}{\mathrm{cos\; 55°}}$ = $\frac{\mathrm{sin\; 25°}}{\mathrm{sin\; 25°}}$ + $\frac{\mathrm{cos\; 55°}}{\mathrm{cos\; 55°}}$

$\frac{\mathrm{cos\; 65°}}{\mathrm{sin\; 25°}}$ + $\frac{\mathrm{sin\; 35°}}{\mathrm{cos\; 55°}}$ = 1 + 1 = 2

Let h = height at which the ladder touches the building.

The angle the ladder makes with the ground is the same for both ladders.

The length of the ladder is the hypotenuse.

sin(θ) = $\frac{\mathrm{Opposite}}{\mathrm{Hypothenuse}}$

10 cm long ladder

Opposite side to θ = 8 cm

Hypothenuse = 10 cm

sin(θ) = $\frac{\mathrm{8\; cm}}{\mathrm{10\; cm}}$ = $\frac{4}{5}$

12 cm long ladder

Opposite side to θ = h cm

Hypothenuse = 12 cm

sin(θ) = $\frac{\mathrm{h\; cm}}{\mathrm{12\; cm}}$ = $\frac{\mathrm{h}}{12}$

Since the angles are the same, the sin for both cases would be the same.

$\frac{\mathrm{h}}{12}$ = $\frac{4}{5}$

Cross multiply.

h x 5 = 12 x 4

Divide both sides by 5

h = $\frac{\mathrm{12\; x\; 4}}{5}$ = 9.6 cm

cos (3x + 28°) = sin (2x + 48°)

Use the trigonometric values to change either cos to sin or sin to cos and equate the angles to solve for x.

Method I

Changing the cos to sin.

cos (3x + 28°) = sin (90° - [3x + 28°])

cos (3x + 28°) = sin (2x + 48°)

sin (90° - [3x + 28°]) = sin (2x + 48°)

Since both sides are sin, the angles are the same.

90° - (3x + 28°) = 2x + 48°

90° - 3x - 28° = 2x + 48°

90° - 28° - 48° = 2x + 3x

90° - 76° = 5x

14° = 5x

Divide both sides by 5.

x = $\frac{\mathrm{14°}}{5}$ = 2.8°

Method II

Changing the sin to cos.

sin (2x + 48°) = cos(90° - [2x + 48°])

cos (3x + 28°) = sin (2x + 48°)

cos (3x + 28°) = cos(90° - [2x + 48°])

Since both sides are cos, the angles are the same.

3x + 28° = 90° - (2x + 48°)

3x + 28° = 90° - 2x - 48°

3x + 2x = 90° - 48° - 28°

5x = 14°

Divide both sides by 5.

x = $\frac{\mathrm{14°}}{5}$ = 2.8°