Answer:
[tex] \angle NRQ =85 \degree[/tex]
Given:
[tex] \angle RPQ = 45 \degree \\ \angle PQR = (6x + 4)\degree \\ \angle NRQ = (15x - 5)\degree [/tex]
Step-by-step explanation:
Property used: An exterior angle of a triangle is equal to the sum of the opposite interior angles.
[tex] = > \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 = (\frac{54}{9} )\degree \\ \\ = > x\degree = 6\degree \\ \\ Putting \: value \: of \: x \: in \: \angle NRQ \\ \\ = > \angle NRQ = (15x - 5)\degree \\ \\ = > \angle NRQ = (15 \times 6 - 5) \degree \\ \\ = > \angle NRQ =85 \degree[/tex]
