Question

Help solve for “q” —————————————

Answer 1

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}

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

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

$\dashrightarrow \tt4q - 1{}^{ \circ}=63^{ \circ}$

$\dashrightarrow \tt4q =63^{ \circ} + 1{}^{ \circ}$

$\dashrightarrow \tt4q =64{}^{ \circ}$

$\dashrightarrow \tt \: q = \dfrac{64}{4}^{ \circ}$

$\dashrightarrow \tt \: q = \dfrac{16 \times 4}{4}^{ \circ}$

$\dashrightarrow \tt \: q = \dfrac{16 \times \cancel4}{\cancel4}^{ \circ}$

$\dashrightarrow \tt \: q = \dfrac{16}{1}$

$\dashrightarrow \bf q = 16 \degree$

Verification:

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

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

$\dashrightarrow \tt63^{ \circ} + {117}^{ \circ} = 18 {0}^{ \circ}$

$\dashrightarrow \tt180^{ \circ} = 18 {0}^{ \circ}$

LHS = RHS

HENCE VERIFIED!