Question

in triangle ABC angle a=x°angle bis (3x-2)°cis y°.cand b=90°.find 3 angles​

Answer 1

Given:-

☆Three angles of a triangle are:-

•x° (<A)

•(3x-2)° (<B)

•90° (<C)

To find:-

☆3 angles of the triangle

Solution:-

We know that,sum of 3 angles of a triangle is 180°

Thus----

=>x+3x-2+90=180

=>4x+88=180

=>4x=180-88

=>4x=92

=>x=92/4

=>x=23

Thus,angles of the triangle are:-

•23° (<A)

•(23×3-2)=67° (<B)

•90° (<C)

Answer 2

Step-by-step explanation:

Given:-

• ☆Three angles of a triangle are:-

•x° (<A)

•(3x-2)° (<B)

•90° (<C)

To find:-

☆3 angles of the triangle

Solution:-

• We know that,sum of 3 angles of a triangle is 180°

Thus----

=>x+3x-2+90=180

=>4x+88=180

=>4x=180-88

=>4x=92

=>x=92/4

=>x=23

• Thus,angles of the triangle are:-

•23° (<A)

•(23×3-2)=67° (<B)

•90° (<C)

hope this helps you!!

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