Solution:

In the right angled ΔABC, AB = 6cm, tanA = $$\frac{4}{3}$$

Now,

or, tanA = $$\frac{BC}{AB}$$

or, $$\frac{4}{3}$$ = $$\frac{BC}{6}$$

or, 3BC = 24

or, BC = $$\frac{24}{3}$$ = 8cm

Also,

AC = $$\sqrt{AB^2+BC^2}$$

= $$\sqrt{6^2+8^2}$$

= $$\sqrt{36+64}$$

= $$\sqrt{100}$$

= 10

∴ AC = 10cm

Solution:

Here, PO = 10cm, OQ = $$\sqrt{2}$$ and ∠O = 90°.
By Pythagoras theorem, we have,
or, PQ2 = PO2 + OQ2
or, PQ2 =102 + ($$\sqrt{8}$$)2
or, PQ2= 100 + 8
or, PQ = $$\sqrt{108}$$
or, PQ = $$\sqrt{36 × 3}$$
or, PQ = $$\sqrt{6^2 × 3}$$
or, PQ = 6 $$\sqrt{3}$$
∴ PQ = 6 $$\sqrt{3}$$ cm

Solution:

Given,
∠B = 90°, ∠A = 30° and c = 5cm
or, ∠C = ∠B − ∠A
or, ∠C = 90°−30°
or, ∠C = 60°
Now,
or, sin60° = $$\frac{p}{h}$$
or, $$\frac{\sqrt{3}}{2}$$ = $$\frac{5}{h}$$
or, $$\sqrt{3}$$ h = 10
∴h = $$\frac{10}{\sqrt{3}}$$
Again,
or, tan60° = $$\frac{p}{b}$$
or, $$\sqrt{3}$$ = $$\frac{5}{b}$$
or, $$\sqrt{3}$$ b = 5
∴ b = $$\frac{5}{\sqrt{3}}$$

Solution:

Given,
∠B = 90°, c = 2cm and a = 2cm
Taking reference angle C
or, tanC = $$\frac{AB}{BC}$$
or, tanC = $$\frac{2}{2}$$
or, tanC = 1
or, tanC = tan45°
∴ C = 45°
Now,
or, ∠A = 90°− 45°
∴ ∠A = 45°
Again,
or, sin45° = $$\frac{p}{h}$$
or, $$\frac{1}{\sqrt{2}}$$ = $$\frac{BC}{AC}$$
or, $$\frac{1}{\sqrt{2}}$$ = $$\frac{2}{AC}$$
∴ AC = 2 $$\sqrt{2}$$cm

Solution:

In right angle triangle ABC
∠A = 90°, b = 6$$\sqrt{3}$$cm, and c = 6 cm
Now,
or, tanB = $$\frac{AC}{AB}$$
or, tanB = $$\frac{6 \sqrt{3}}{6}$$
or, tanB = $$\sqrt{3}$$
or, tanB = tan60°
∴ b = 60°

Taking reference∠C
or, sinC = $$\frac{AB}{BC}$$
or, sin30° = $$\frac{6}{BC}$$
or, $$\frac{1}{2}$$ = $$\frac{6}{BC}$$
∴ BC = 12cm.

Solution:

In a right angle triangle
∠B = 90°, a = $$\sqrt{3}$$ cm and c = 1cm
Taking reference angle C
or, tanC = $$\frac{AB}{BC}$$
or, tanC = $$\frac{1}{\sqrt{3}}$$
or, tanC = tan30°
∴ C = 60°
Now,
or, ∠A = 90°− 30°
or, ∠A = 60°

Taking reference angle A
or, sinA = $$\frac{BC}{AC}$$
or, sin60° =$$\frac{BC}{AC}$$
or, $$\frac{\sqrt{3}}{2}$$ = $$\frac{\sqrt{3}}{AC}$$
or, $$\sqrt{3}$$ AC = 2$$\sqrt{2}$$
or, AC = $$\frac{2\sqrt{3}}{\sqrt{3}}$$
∴ AC = 2cm

Solution:

In right angle
∠C = 90°, ∠A = 30° and b = 20 cm
Taking∠A as a reference
or, ∠B = ∠C−∠A
or, ∠B = 90° − 30°
or, ∠B = 60°

Taking ∠B as a reference
or, sinB = $$\frac{p}{h}$$
or, sin60° = $$\frac{20}{C}$$
or, $$\frac{\sqrt{3}}{2}$$ = $$\frac{20}{C}$$
or, $$\sqrt{3}$$ C = 40
or, C = $$\frac{40}{\sqrt{3}}$$
Then,
or, cos60° = $$\frac{b}{h}$$
or, $$\frac{1}{2}$$ = $$\frac{a}{\frac{40}{\sqrt{3}}}$$
or, $$\frac{1}{2}$$ = $$\frac{a× \sqrt{3}}{40}$$
or, 40 = 2$$\sqrt{3}$$ a
or, a = $$\frac{40}{2\sqrt{3}}$$
∴ a = $$\frac{20}{\sqrt{3}}$$

Solution:

In right angle triangle $$\angle$$PQR,$$\angle$$Q = 90o, $$\angle$$R = $$\theta$$ , PQ = 5cm, QR = 12cm

Now,

Hypotenuse(h) = ?

Perpendicular(p) = 5cm

Base(b) = 12cm

We know, h = $$\sqrt{b^2-p^2}$$

=$$\sqrt{12^2-5^2}$$

=$$\sqrt{144-25}$$

=$$\sqrt{169}$$

= 13cm

Now,

sin$$\theta$$ = $$\frac{p}{h}$$ = $$\frac{5}{13}$$

cos$$\theta$$ =$$\frac{b}{h}$$ =$$\frac{12}{13}$$

tan$$\theta$$ =$$\frac{p}{b}$$ =$$\frac{5}{12}$$

cosec$$\theta$$ =$$\frac{h}{p}$$ =$$\frac{13}{5}$$

sec$$\theta$$ =$$\frac{h}{b}$$ =$$\frac{5}{13}$$

cot$$\theta$$ =$$\frac{b}{p}$$ =$$\frac{12}{5}$$

Solution:

Hypotenuse(h) = 20cm

Perpendicular(p) = 16cm

Base(b) = 12cm

We know,

h2= p2+ b2

or, 202= 162+ 122

or, 400 = 256 + 144

$$\therefore$$ 400 = 400 (It satisfy pythagorus theorem)

Hence, the given triangle is right angled triangle.

Solution:

Here,

AB, BC and CA are perpendicular, base and hypotenuse respectively.

Now,

∴ sinθ = $$\frac{p}{h}$$ = $$\frac{AB}{AC}$$

∴ cosθ = $$\frac{b}{h}$$ = $$\frac{BC}{AC}$$

∴ tanθ = $$\frac{p}{b}$$ = $$\frac{AB}{BC}$$

Solution:

In right angle triangle ABC, AC = 10cm

sinA = $$\frac{BC}{AC}$$

or, $$\frac{4}{5}$$ = $$\frac{BC}{10}$$

or, 5BC = 10 x 4

or, BC = $$\frac{40}{5}$$

$$\therefore$$ BC = 8 cm

By using pythagorus theorum

or, AB2= AC2 - BC2

or, AB = $$\sqrt{(10^2 - 8^2)}$$

or, AB = $$\sqrt{100 - 64}$$

or, AB = $$\sqrt{36}$$

$$\therefore AB = 6cm Hence, the length of AB is 6cm. Solution: Here, ∠O = 90°, ∠N = 30° and n = 2\(\sqrt{3}$$cm
NO = ? and NM = ?

From the right angledΔMNO, we have
or, ∠N + ∠M = 90°
or, 30° + ∠M = 90°
or, ∠M = 90°− 30°
or, ∠M = 60°
Now,
or, tan30° = $$\frac{OM}{NO}$$
or, $$\frac{1}{\sqrt{3}}$$ = $$\frac{2\sqrt{3}}{m}$$
or, m = 2× 3
∴ m = 6cm
Again,
or, sin30° = $$\frac{OM}{NM}$$
or, $$\frac{1}{2}$$ = $$\frac{2\sqrt{3}}{o}$$
or, o = 4$$\sqrt{3}$$
Hence, o = 4$$\sqrt{3}$$, m = 6cm and ∠M = 60°

Solution:

Given,
∠B = 90°, ∠C = 60° and AB = 3cm
Now,
or, ∠A = ∠B − ∠C
or, ∠A = 90°− 60°
∴ ∠A = 30°
Again,
or, sin60° = $$\frac{AB}{AC}$$
or, $$\frac{\sqrt{3}}{2}$$ = $$\frac{3}{AC}$$
or, $$\sqrt{3}{AC}$$ = 3 × 2
or, AC = $$\frac{3 × 2}{\sqrt{3}}$$
or, AC = $$\frac{\sqrt{3}× \sqrt{3} × 2}{\sqrt{3}}$$
∴ AC = 2$$\sqrt{3}$$
Then.
or, sin30° = $$\frac{p}{h}$$
or, $$\frac{1}{2}$$ = $$\frac{BC}{2\sqrt{3}}$$
or, 2 BC = 2$$\sqrt{3}$$
or, BC = $$\frac{2\sqrt{3}}{2}$$
∴ BC = $$\sqrt{3}$$
∴ The value of AC is 2$$\sqrt{3}$$ and BC is $$\sqrt{3}$$.