Find the value of x if interior angles are x°, 3y° and 5y° and exterior angle of a triangle is 7y°
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$\Delta ABC$ is triangle and its interior angles are $x^\circ$, $3y^\circ$ and $5y^\circ$. The side $\overrightarrow{AB}$ of the triangle is extended through a point $D$. Thus, the triangle $ABC$ has an exterior angle $CBD$ and it is $7y^\circ$.
In this geometry problem, the literals $x$ and $y$ are used to represent angles but they are symbolically written as $x^\circ$ and $y^\circ$. The meaning of representing $x$ and $y$ as $x^\circ$ and $y^\circ$ is the values of $x$ and $y$ are in degrees, which means the literal numbers $x$ and $y$ represent angles in degree measuring system.
Now, let us find the value of $x$ by using the geometrical relation between $y$ and $x$.
Step: 1Forget about the triangle $ABC$ but only consider the intersecting lines $\overrightarrow{AD}$ and $\overline{BC}$. Geometrically, the line segment $\overline{BC}$ divides the angle $ABD$ into two parts and the angles are $5y$ and $7y$ in degrees.
Therefore $\angle CBA = 5y$ and $\angle CBD = 7y$ in degrees.
The line $\overrightarrow{AD}$ is a straight ray. So, the angle of the line $\overrightarrow{AD}$ is $\angle ABD$ and it is a straight angle. Therefore, the angle $ABD$ is $180^\circ$ geometrically.
Geometrically, the sum of the angles $\angle CBA$ and $\angle CBD$ is equal to $\angle ABD$.
$\angle CBD + \angle CBA = \angle ABD$
$\implies 7y + 5y = 180^\circ$
$\implies 12y = 180^\circ$
$\implies y = \dfrac{180^\circ}{12}$
$\implies \require{cancel} y = \dfrac{\cancel{180^\circ}}{\cancel{12}}$
$\therefore \,\,\,\,\,\, y = 15^\circ$
Therefore, the value of $y$ is $15^\circ$.
Step: 2The value of literal $y$ in degrees is calculated in previous step and it is $15^\circ$. Now, consider only $\Delta ABC$. Geometrically, it is proved that the sum of three angles of a triangle is $180^\circ$.
It can be expressed in mathematical form as follows.
$\angle CBA + \angle BAC + \angle ACB = 180^\circ$
Substitute the values of all three angles of triangle $ABC$ in algebraic form.
$\implies 5y + x + 3y = 180^\circ$
$\implies x + 5y + 3y = 180^\circ$
$\implies x + 8y = 180^\circ$
$\implies x = 180^\circ -8y$
$\implies x = 180^\circ -8 (15^\circ)$
$\implies x = 180^\circ -120^\circ$
$\therefore \,\,\,\,\,\, x = 60^\circ$
Therefore, the value of $x$ is $60^\circ$ and it is the required solution for this geometric problem.
Solution 1:
(i) \angle A\ +\ \angle B+\ \angle C\ =\ 180\degree { angle sum property of triangle }
70\degree+\ 50\degree+\ \angle B\ =\ 180\degree
\angle B\ =\ 180\degree-\ 120\degree
\angle B\ =\ 60\degree
(ii) \angle BAC\ +\angle ACB\ =\ x\ {an exterior angle of a triangle is equal to the sum of the opposite interior angle }
65\degree+45\degree=x
110\degree=x
(iii) \angle CAB+\angle ABC\ =\ x {an exterior angle of a triangle is equal to the sum of the opposite interior angle }
30\degree+40\degree\ =\ x\
70\degree=x
(iv) \angle BCA+\angle CBA\ =\ x {an exterior angle of a triangle is equal to the sum of the opposite interior angle }
60\degree+60\degree\ =\ x
120\degree=x
(v) \angle A+\angle B=\ x {an exterior angle of a triangle is equal to the sum of the opposite interior angle }
50\degree+50\degree=\ x
100\degree=\ x\
(vi) \angle A+\angle B\ =\ x\ {an exterior angle of a triangle is equal to the sum of the opposite interior angle }
30\degree+60\degree\ =\ x\
90\degree=\ x
Properties of Triangles
Question 1: Find the value of unknown exterior angle in following diagrams.
Answer:
(i) `x = 50° + 70° = 120°`
(ii) `x = 65° + 45° = 110°`
(iii) `x = 30° + 40° = 70°`
(iv) `x = 60° + 60° = 120°`
(v) `x = 50° + 50° = 100°`
(vi) `x = 60° + 30° = 90°`
Chapter List
Integers Fraction & Decimal Data Handling Simple Equation Lines & Angles Triangles Congruence of Triangles Comparing Quantities Rational Numbers Practical Geometry Area Perimeter Algebra Exponents & Power Symmetry Solid ShapesQuestion 2: Find the value of unknown interior angle in following figures:
Answer: Exterior angle of a triangle is equal to the sum of opposite interior angles.
(i) `x = 115° - 50° = 65°`
(ii) `x = 100° - 70° = 30°`
(iii) `x = 120° - 60° = 60°`
(iv) `x = 80° - 30° = 50°`
(v) `x = 75° - 35° = 40°`
Exercise 6.3
Question 1: Find the value of unknown x in following figures:
(a) Answer: `x = 180° - (50° + 60°) ``= 180° - 110° = 70°`
(b) Answer: `x = 180° - (30° + 90°)`
Since it is a right angle so the third angle is a right angle.
Or, `x = 180° - 120° = 60°`
(c) Answer: `x = 180° - (30° + 110°) ``= 180° - 140° = 40°`
(d) Answer: Here; `50° + 2x = 180°`
Or, `2x = 180° - 50° = 130°`
Or, `x = 130° ÷ 2 = 65°`
(e) Answer: This is an equilateral triangle
Hence, `3x = 180°`
Or, `x = 180° ÷ 3 = 60°`
(f) Answer: This is a right angled triangle.
Hence, `2x + x + 90°= 180°`
Or, `3x = 180° - 90°`
Or, `x = 90° ÷ 3 = 30°`
Question 2: Find the values of the unknowns x and y in the following diagrams.
(i) Answer: Since an external angle is equal to the sum of opposite exterior angles.
Hence, `120° = 50° + x`
Or, `x = 120° - 50° = 70°`
Now, `120° + y = 180°`
Because they make linear pair of angles and angles of a linear pair are always supplementary.
Or, `y = 180° - 120° = 60°`
(ii) Answer: In this case, y = 80°
Because, vertically opposite angles are always equal.
Now, `50° + y + x = 180°` (Angle sum property of triangle)
Or, `50° + 80° + x = 180°`
Or, `x + 130° = 180°`
Or, `x = 180° - 130°= 50°`
(iii) Answer: Here, x = 50° + 60° = 110°
Because , exterior angle in a triangle is equal to sum of opposite internal angles.
Now, `x + y = 180°` (Linear pair of angles are supplementary)
Or, `110° + y = 180°`
Or, `y = 180° - 110° = 70°`
(iv) Answer: Here, `x = 60°` (Vertically opposite angles are equal)
Now, `30° + x + y = 180°` (Angle sum of triangle)
Or, `30° + 60° + y = 180°`
Or, `y + 90° = 180°`
Or, `y = 180° - 90°`
Or, `x = 60°` and `y = 90°`
(v) Answer: Here, x = 90° (Vertically opposite angles are equal)
Now, `x + x + y = 180°`
Or, `2x + 90° = 180°`
Or, `2x = 180° - 90° = 90°`
Or, `x = 90° ÷ 2 = 45°`
(vi) Answer: Here, x = y (Vertically opposite angles are equal.
Thus, all angles of the given triangle are equal. It means that the given triangle is equilateral triangle and each angle has same measure.
Hence, `x = y = 60°`
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