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leonid [27]
3 years ago
5

Can someone help me with this​

Mathematics
1 answer:
SashulF [63]3 years ago
4 0

Answer:

\angle NRQ =85 \degree

Given:

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

Step-by-step explanation:

Property used: <em>An exterior angle of a triangle is equal to the sum of the opposite int</em><em>e</em><em>rior angles.</em>

<em>=  >  \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</em>

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