Question

plzz solve the 5 and 6 que in long method

Answer 1

Q.5) since, triangle is isosceles,

so, angle 1 = 32° (given)

also, angle 2 = 32° (given)

since, sum of all angles of a triangle is 180°

then, by angle sum property,

angle 1 + angle 2 + angle 3 = 180°

=> 32° + 32° + angle 3 = 180°

=> 64° + angle 3 = 180°

=> angle 3 = 180° - 64°

=> angle 3 = 116° ......answer
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Q.6)

let x be the measure of an angle whose supplement is equal to angle itself.

we known that, sum of two supplementary angles is 180°

so,

x + x = 180°

=> 2x = 180°

=> x = $\frac{180}{2}$

=> x = 90° .....answer.

Answer 2

Here is your solution


[Q.5]

Given:-

Triangle is isosceles

∠1 = 32° (given)

∠2 = 32° (given)

we know that

sum of all angles of a triangle is 180°

Now,

∠1 + ∠2 + ∠ 3 = 180°

=> 32° + 32° + ∠3 = 180°

=> 64° + ∠3 = 180°

=> ∠3 = 180° - 64°

=> ∠3 = 116° ✔

[Q.6]

Let,

x be the measure of an angle whose supplement is equal to angle itself.

we known that, sum of two supplementary angles is 180°

Now

A/q

=>x + x = 180° 

=> 2x = 180°

$x = \frac{180}{2}$

=> x = 90° ✔