Solution of Right Angled Triangle
Solution of Right Angled Triangle
The triangle which consists three sides and three angles with six elements is known as right angled triangle. In right angle triangle one angle is 90. If three elements are given, one of which must be the side and remaining others elements can be calculated which is known as solution of right angle triangle.
Let,
In a given figure, right angled triangle ABC, right angled at B, ∠A, ∠B and ∠C represent the angles and a, b and c represent their opposite side. Then ΔABC, using pythagoras theorem,
AC^{2} = AB^{2} + BC^{2}
i.e, b^{2}= c^{2}+ a^{2}
and ∠A + ∠C = 90°
 The triangle which consists three sides and three angles with six elements is known as right angled triangle.
 In right angle triangle one angle is 90.
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, PQ^{2} = PO^{2} + OQ^{2}
or, PQ^{2 }=10^{2} + (\(\sqrt{8}\))^{2}or, PQ^{2}= 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 = 90^{o}, \(\angle\)R = \(\theta\) , PQ = 5cm, QR = 12cm
Now,
Hypotenuse(h) = ?
Perpendicular(p) = 5cm
Base(b) = 12cm
We know, h = \(\sqrt{b^2p^2}\)
=\(\sqrt{12^25^2}\)
=\(\sqrt{14425}\)
=\(\sqrt{169}\)
= 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,
h^{2}= p^{2}+ b^{2}
or, 20^{2}= 16^{2}+ 12^{2}
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, AB^{2}= AC^{2}  BC^{2}
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 \(\sqrt{3}\).

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bhupendrahow to find remaining parts of right angled triangle, if <b=90°,, b=6cm,a=4cm 
Feb 08, 2017 
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gaurinaA man on the top of a cliff 90m high observes a boat on the river and finds the angle of depression to be 60°. How far is the boat from the cliff? 
Jan 03, 2017 
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