Question

Help solve for “q”
—————————————

Answer 1

Value of
`\sf\purple{q\: = \:16.}`

Explanation:

`\rightarrow`As we know that,

Sum all angles that lie on a straight line =
`\sf\blue{180°}`

So,

`\rightarrow`
`\sf{(4q-1)°+ 117°\: = \:180°}`

`\rightarrow`
`\sf{(4q-1)\: = \:180-117}`

`\rightarrow`
`\sf{(4q-1)\: = \:63}`

`\rightarrow`
`\sf{4q\: = \:63+1}`

`\rightarrow`
`\sf{q\: = \:(64)/(4)}`

`\rightarrow`
`\sf{q\: = \:16}`

Thus,
`\sf\purple{q\: = \:16.}`

_________________________________

Hope it helps you:)

Answer 2

Digram:-

`\\`

`\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\put(5,1){\vector(1,0){4}}\put(5,1){\vector(-1,0){4}}\put(5,1){\vector(1,1){3}}\put(2,2){$\underline{\boxed{\large\sf a + b = 180^(\circ)}$}}\put(4.5,1.3){$\sf a^(\circ)$}\put(5.7,1.3){$\sf b^(\circ)$}\end{picture}`

`\\`

STEP :-

`\dashrightarrow \tt(4q - 1) {}^( \circ) + {117}^( \circ) = 18 {0}^( \circ)`

{Linear pair}

`\\ \\`

`\dashrightarrow \tt(4q - 1) {}^( \circ)= 18 {0}^( \circ) - {117}^( \circ)`

`\\`

`\dashrightarrow \tt(4q - 1) {}^( \circ)=63^( \circ)`

`\\`

`\dashrightarrow \tt4q - 1{}^( \circ)=63^( \circ)`

`\\`

`\dashrightarrow \tt4q =63^( \circ) + 1{}^( \circ)`

`\\`

`\dashrightarrow \tt4q =64{}^( \circ)`

`\\`

`\dashrightarrow \tt \: q = (64)/(4)^( \circ)`

`\\`

`\dashrightarrow \tt \: q = (16 * 4)/(4)^( \circ)`

`\\`

`\dashrightarrow \tt \: q = (16 * \cancel4)/(\cancel4)^( \circ)`

`\\`

`\dashrightarrow \tt \: q = (16)/(1)`

`\\`

`\dashrightarrow \bf q = 16 \degree`

`\\ \\`

Verification:

`\\`

`\dashrightarrow \tt(4 * 16- 1) {}^( \circ) + {117}^( \circ) = 18 {0}^( \circ)`

`\\`

`\dashrightarrow \tt(64- 1) {}^( \circ) + {117}^( \circ) = 18 {0}^( \circ)`

`\\`

`\dashrightarrow \tt63^( \circ) + {117}^( \circ) = 18 {0}^( \circ)`

`\\`

`\dashrightarrow \tt180^( \circ) = 18 {0}^( \circ)`

`\\`

LHS = RHS

HENCE VERIFIED!