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pashok25 [27]
3 years ago
7

Find the missing length. The triangles are similar. Can someone help me?

Mathematics
1 answer:
kirill [66]3 years ago
8 0

Answer:

The length of QS = 14 _________

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4 years ago
Help solve for “q”<br> —————————————
VMariaS [17]

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!

8 0
3 years ago
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