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PLEASE HELP !!!!! ASAP PLEASE !!!! IM BEGGING LOL !!!!

PLEASE HELP ASAP PLEASE IM BEGGING LOL class=

Sagot :

Sueraiuka

Step-by-step explanation:

Hey there!

From the given figure;

Angle ABE = 90°.

And angle ABC= [(3x+7)° + (5x+11)°].

Now, Angle ABE + Angle ABC= 180° [Being linear pair].

or, 90° + (3x+7)° + (5x+11)°= 180°

or, (3x+7)° + (5x+11)°= 90°

or, 8x + 18° =90°

or, 8x = 90°-18°

or, X = 9°

Therefore, X= 9°.

Now,

Angle DBC = (5*9+11)°

= 56°

Therefore, the measure of angle DBC is 56°.

Hope it helps!

[tex]\huge\bold{To\:find:}[/tex]

✎ The measure of ∠ DBC.

[tex]\large\mathfrak{{\pmb{\underline{\orange{Solution}}{\orange{:}}}}}[/tex]

[tex]\sf\purple{The\:measure\:of\:∠DBC\:is\:56°. }[/tex]

[tex]\large\mathfrak{{\pmb{\underline{\red{Step-by-step\:explanation}}{\orange{:}}}}}[/tex]

Since it is right-angle,

[tex]Sum \: of \: angles \: = 90° \\ ⇢ (3x + 7)° + (5x + 11) ° = 90° \\ ⇢3x° + 7° + 5x° + 11° = 90° \\combining \: the \: like \: terms, \: we \: have \\ ⇢3x° + 5x° + 7° + 11° = 90° \\ ⇢8x° = 90° - 18° \\ ⇢8x° = 72° \\ ⇢x° = \frac{72°}{8} \\ ⇢x° = 9°[/tex]

Substituting the value of [tex]x°[/tex] in ∠ DBC, we have

[tex](5x + 11) °\\ = 5 \times 9° + 11 °\\ = 45° + 11° \\ = 56°[/tex]

[tex]\sf\blue{Therefore,\:the\:measure\:of\:∠DBC\:is\:56°.}[/tex]

[tex]\huge\bold{To\:verify :}[/tex]

[tex](3x + 7) ° + (5x + 11)° = 90° \\ \: ➡ \: (3 \times 9 + 7)° + (5 \times 9 + 11)° = 90° \\ ➡ \: 34° + 56° = 90° \\ ➡ \: 90° = 90° \\ ➡ \: L.H.S.=R. H. S[/tex]

Hence verified.

[tex]\circ \: \: { \underline{ \boxed{ \sf{ \color{green}{Happy\:learning.}}}}}∘[/tex]

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