Question

Can someone help me with this​

Answer 1

`\angle NRQ =85 \degree`

Given:

`\angle RPQ = 45 \degree \\ \angle PQR = (6x + 4)\degree \\ \angle NRQ = (15x - 5)\degree`

Explanation:

Property used: An exterior angle of a triangle is equal to the sum of the opposite interior angles.

`= > \angle RPQ + \angle PQR = \angle NRQ \\ \\ = > 45 \degree + (6x + 4)\degree = (15x - 5)\degree \\ \\ = > 45\degree + 6x\degree + 4\degree = 15x\degree - 5\degree \\ \\ = > 49\degree + 5 \degree= 15x\degree - 6x\degree \\ \\ = > 54\degree = 9x\degree \\ \\ = > 9x\degree = 54\degree \\ \\ = > x\degree = ((54)/(9) )\degree \\ \\ = > x\degree = 6\degree \\ \</em><em>\</em><em> </em><em>Putting</em><em> </em><em>\</em><em>:</em><em> </em>value<em> </em><em>\</em><em>:</em><em> </em>of<em> </em><em>\</em><em>:</em> x \: in \: \angle NRQ <em> \\ \\ = > \angle NRQ = (15x - 5)\degree \\ \\ = > \angle NRQ = (15 * 6 - 5) \degree \\ \\ = > \angle NRQ =85 \degree`