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Find the common external angle of two triangles, if interior angles of one triangle are 25° and 55° and of other triangle is 40°

$\Delta ABD$ and $\Delta CBE$ are two triangles. The triangle $ABD$ has two known angles and they are $25^\circ$ and $55^\circ$ but third angle is unknown.

triangles with common unknown external angle

The triangle $BCE$ has one known angle but the remaining two angles are unknown.

The two triangles has a common side $\overline{BE}$ and also have an external angle commonly and it is denoted by $x$.

In this problem, it is required to find the value of $x$ by studying the geometrical relations between angles of these two triangles.

Step: 1

In $\Delta ABD$, two of three angles are known and they are $25^\circ$ and $55^\circ$ but third angle is unknown. However, it can be found by applying the sum of three angles of a triangle rule.

third angle of the first triangle

$\angle DAB + \angle ABD + \angle BDA$ $=$ $180^\circ$

The angle $DAB$ is $55^\circ$ and the angle $BDA$ is $25^\circ$. Substitute them in the equation to obtain the angle $ABD$.

$\implies 55^\circ + \angle ABD + 25^\circ$ $=$ $180^\circ$

$\implies \angle ABD + 80^\circ$ $=$ $180^\circ$

$\implies \angle ABD$ $=$ $180^\circ -80^\circ$

$\therefore \,\,\,\,\,\, \angle ABD$ $=$ $100^\circ$

It is derived that the angle $ABD$ is $100^\circ$ in the triangle $ABD$.

Step: 2

In triangle $BCE$, only one angle is known and it is $40^\circ$ but the two angles are unknown. If one of two angles is found, then third angle of the triangle can be easily evaluated.

straight angle between triangles

Now, consider the intersecting lines $\overline{AC}$ and $\overline{BD}$. Actually, $\overline{AC}$ is a straight line and its angle is $180^\circ$ geometrically. The line segment $\overline{BD}$ intersects the straight line $\overline{AC}$ at point $B$. So, it splits the angle $180^\circ$ into two sub angles and they are $\angle ABD$ and $\angle DBC$ but the angle $\angle ABD$ is $100^\circ$ as per previous step.

$\angle ABD + \angle DBC = 180^\circ$

$\implies 100^\circ + \angle DBC = 180^\circ$

$\implies \angle DBC = 180^\circ -100^\circ$

$\therefore \,\,\,\,\,\, \angle DBC = 80^\circ$

Therefore, it is derived that the angle $DBC$ is $80^\circ$ and it is the second angle of the $\Delta BCE$.

Step: 3

Now, the two interior angles are known for the triangle $BCE$. Apply sum of angles of the triangle rule once again to find the third angle of the triangle $BCE$.

angle of the triangle

$\angle EBC + \angle BCE + \angle CEB = 180^\circ$

$\implies 80^\circ + 40^\circ + \angle CEB = 180^\circ$

$\implies 120^\circ + \angle CEB = 180^\circ$

$\implies \angle CEB = 180^\circ -120^\circ$

$\therefore \,\,\,\,\,\, \angle CEB = 60^\circ$

The third angle of the triangle $BCE$ has been evaluated in this step.

Step: 4
common externor angle of the triangles

$\overline{BD}$ is a straight line and its angle is $180^\circ$ but the angle is divided by the line segment $\overline{CE}$ into two parts. One of the sub angle is $\angle CEB$ and it is evaluated as $60^\circ$. So, the angle $CED$ can be evaluated geometrically.

$\angle CEB + \angle CED = \angle BED$

$\implies 60^\circ + x = 180^\circ$

$\implies x = 180^\circ -60^\circ$

$\therefore \,\,\,\,\,\, x = 120^\circ$

The value of $x$ is $120^\circ$ and it is the required solution for this geometrical problem in mathematics.

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