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 \(\sqrt{3}\).