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Help solve for “q”
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Help Solve For Q class=

Sagot :

Digram:-

[tex] \\ [/tex]

[tex]\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}[/tex]

[tex] \\ [/tex]

STEP :-

[tex] \dashrightarrow \tt(4q - 1) {}^{ \circ} + {117}^{ \circ} = 18 {0}^{ \circ} [/tex]

{Linear pair}

[tex] \\ \\ [/tex]

[tex] \dashrightarrow \tt(4q - 1) {}^{ \circ}= 18 {0}^{ \circ} - {117}^{ \circ}[/tex]

[tex] \\ [/tex]

[tex] \dashrightarrow \tt(4q - 1) {}^{ \circ}=63^{ \circ}[/tex]

[tex] \\ [/tex]

[tex] \dashrightarrow \tt4q - 1{}^{ \circ}=63^{ \circ}[/tex]

[tex] \\ [/tex]

[tex] \dashrightarrow \tt4q =63^{ \circ} + 1{}^{ \circ}[/tex]

[tex] \\ [/tex]

[tex] \dashrightarrow \tt4q =64{}^{ \circ}[/tex]

[tex] \\ [/tex]

[tex] \dashrightarrow \tt \: q = \dfrac{64}{4}^{ \circ}[/tex]

[tex] \\ [/tex]

[tex] \dashrightarrow \tt \: q = \dfrac{16 \times 4}{4}^{ \circ}[/tex]

[tex] \\ [/tex]

[tex] \dashrightarrow \tt \: q = \dfrac{16 \times \cancel4}{\cancel4}^{ \circ}[/tex]

[tex] \\ [/tex]

[tex] \dashrightarrow \tt \: q = \dfrac{16}{1}[/tex]

[tex] \\ [/tex]

[tex] \dashrightarrow \bf q = 16 \degree[/tex]

[tex] \\ \\ [/tex]

Verification:

[tex] \\ [/tex]

[tex] \dashrightarrow \tt(4 \times 16- 1) {}^{ \circ} + {117}^{ \circ} = 18 {0}^{ \circ} [/tex]

[tex] \\ [/tex]

[tex] \dashrightarrow \tt(64- 1) {}^{ \circ} + {117}^{ \circ} = 18 {0}^{ \circ} [/tex]

[tex] \\ [/tex]

[tex] \dashrightarrow \tt63^{ \circ} + {117}^{ \circ} = 18 {0}^{ \circ} [/tex]

[tex] \\ [/tex]

[tex] \dashrightarrow \tt180^{ \circ} = 18 {0}^{ \circ} [/tex]

[tex] \\ [/tex]

LHS = RHS

HENCE VERIFIED!

Answer:

Value of [tex]\sf\purple{q\: = \:16.}[/tex]

Step-by-step explanation:

[tex]\rightarrow[/tex]As we know that,

Sum all angles that lie on a straight line = [tex]\sf\blue{180°}[/tex]

So,

[tex]\rightarrow[/tex] [tex]\sf{(4q-1)°+ 117°\: = \:180°}[/tex]

[tex]\rightarrow[/tex] [tex]\sf{(4q-1)\: = \:180-117}[/tex]

[tex]\rightarrow[/tex] [tex]\sf{(4q-1)\: = \:63}[/tex]

[tex]\rightarrow[/tex] [tex]\sf{4q\: = \:63+1}[/tex]

[tex]\rightarrow[/tex] [tex]\sf{q\: = \:\frac{64}{4}}[/tex]

[tex]\rightarrow[/tex] [tex]\sf{q\: = \:16}[/tex]

Thus, [tex]\sf\purple{q\: = \:16.}[/tex]

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Hope it helps you:)