Question

find the value of x
describe​

Answer 1

Answer:

135-90=40degree

50+30=80degree

Answer 2

$\Large{\underbrace{\sf{\red{Required\:Solution:}}}}$

Sol'n (i). According to the question,

$\longrightarrow{\sf{ x = 30^{\circ} + 50^{\circ}}}$

• Reαson : An exterior αngle of α triαngle is equαl to the sum of the two opposite interior αngles.

$\large\implies{\boxed{\sf{\red{ x = 80^{\circ}}}}}$

Alternαte method :

$\longrightarrow{\sf{ \angle ABC + \angle BCA + \angle CAB = 180^{\circ}}}$

• Reαson : The sum of αll αngles in α triαngle is 180°.

$\longrightarrow{\sf{ 30^{\circ} + 50^{\circ} + \angle CAB = 180^{\circ}}}$

$\longrightarrow{\sf{ 80^{\circ} + \angle CAB = 180^{\circ}}}$

$\longrightarrow{\sf{ \angle CAB = 180^{\circ} - 80^{\circ}}}$

$\longrightarrow{\sf{ \angle CAB = 100^{\circ}}}$

Now,

$\longrightarrow{\sf{ \angle CAB + \angle DAB = 180^{\circ}}}$

• Reαson : Angle sum property of α triαngle.

$\longrightarrow{\sf{ 100^{\circ} + x = 180^{\circ}}}$

$\longrightarrow{\sf{ x = 180^{\circ} - 100^{\circ}}}$

$\large\implies{\boxed{\sf{\red{ x = 80^{\circ}}}}}$

Hence, the vαlue of x is 80°.

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Sol'n (ii). According to the question,

$\longrightarrow{\sf{ 135^{\circ} = 90^{\circ} + x^{\circ}}}$

• Reαson : An exterior αngle of α triαngle is equαl to the sum of the two opposite interior αngles.

$\longrightarrow{\sf{ 135^{\circ} - 90^{\circ} = x}}$

$\large\implies{\boxed{\sf{\red{ x = 45^{\circ}}}}}$

Alternαte method :

$\longrightarrow{\sf{ \angle PQR + \angle QRP + \angle RPQ= 180^{\circ}}}$

• Reαson : The sum of αll αngles in α triαngle is 180°.

$\longrightarrow{\sf{ (180-135)^{\circ} + 90^{\circ} + \angle CAB = 180^{\circ}}}$

$\longrightarrow{\sf{ 45^{\circ} + 90^{\circ}+ \angle CAB = 180^{\circ}}}$

$\longrightarrow{\sf{ 135^{\circ}+ \angle CAB = 180^{\circ}}}$

$\longrightarrow{\sf{ \angle CAB = 180^{\circ} - 135^{\circ}}}$

$\longrightarrow{\sf{ \angle CAB = 45^{\circ}}}$

$\large\implies{\boxed{\sf{\red{ x = 45^{\circ}}}}}$

Hence, the vαlue of x is 45°.

And we αre done! :D

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