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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^{o}. If three elements are given, one of which must be the side and remaining others elements can be calculated which is known as a 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 a Pythagoras theorem,
AC^{2} = AB^{2} + BC^{2}
i.e, b^{2}= c^{2}+ a^{2}
and ∠A + ∠C = 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 = \(\sqrt{b^2-p^2}\)
=\(\sqrt{12^2-5^2}\)
=\(\sqrt{144-25}\)
=\(\sqrt{169}\)
= 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,
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.
ASK ANY QUESTION ON Solution of Right Angled Triangle
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bhupendra
how to find remaining parts of right angled triangle, if <b=90°,, b=6cm,a=4cm
Feb 08, 2017
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gaurina
A 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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