Question

the angles of triangles are 5x-10 degree,2x+40 degree and 3x-50 degree .the triangle is​

Answer 1

Answer:

(angle sum property) sum of all angles of triangle = 180°

(5x-10)+(2x+40)+(3x-50)=180°

5x-10+2x+40+3x-50= 180°

10x-20=180°

10x= 200

x= 20

therefore angles are ( by putting value of x).

5x-10= 5(20)-10= 100-10= 90°

2x+40= 2(20)+40= 40+40= 80°

3x-50= 3(20)-50= 60-50= 10°

check-

90°+80°+10°= 180°

180°=180°

l.h.s = r.h.s

hence proved.

therefore our calculation is correct.

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Answer 2

Concepts Required :

By Angle Sum property of triangle

• Sum of all angle in a triangle be 180° .
• i.e. angle A + angle B + angle C = 180°.

Solution :

Let's Sides of triangle be A, B ,C.

A/Q,

$\tt\:\:\:\dagger\:\:\:\angle A = 5x - 10.$

$\tt\:\:\:\dagger \:\:\:\angle B = 2x + 40.$

$\tt\:\:\:\dagger \:\:\:\angle C = 3x - 50.$

$\rule{90}1$

Now,

By Angle Sum property of Triangle.

$\tt \longmapsto \angle A + \angle B + \angle C =180 \degree$

$\tt\longmapsto (5x - 10) + (2x + 40) + (3x - 50) = 180.$

$\tt \longmapsto 10x + 40 - 60 = 180.$

$\tt\longmapsto 10x - 20 = 180.$

$\tt\longmapsto 10x = 180 + 20.$

$\tt\longmapsto 10x = 200.$

$\tt \longmapsto x = 20 \degree.$

$\rule{200}3$

Now, We have the value of x.

Let find there all sides,

$\tt\mapsto \angle A = 5x - 10$

$\tt \mapsto \angle A=5 (20) - 10.$

$\tt \mapsto \angle A= 100 - 10.$

$\tt\mapsto \angle A = 90.$

$\rule{90}1$

And,

$\tt \mapsto \angle B = 2x + 40.$

$\tt \mapsto \angle B = 2(20) + 40.$

$\tt\mapsto \angle B = 40 + 40$

$\tt \mapsto \angle B = 80.$

$\rule{90}1$

And,

$\tt\mapsto \angle C = 3x - 50.$

$\tt\mapsto \angle C= 3(20) - 50.$

$\tt \mapsto \angle C= 60 - 50$

$\tt\mapsto \angle C = 10.$

$\rule{120}1$

Note : Diagram provide in attachment.