Two angles measure 30-a- and 125-2a- - If each one is the supplement of the other- then the value of a is-

The correct option is C 25^∘ Sum of supplementary angles is 180 ^∘ #160; ∴ (30 a) ^∘ #160; + (125+2a) ^∘ #160; = 180 ^∘ #160; ⇒ a+ 1 ...

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Byju's Answer

Standard VI

Mathematics

Types of Angles

Two angles me...

Question

Two angles measure $(30 - a)^{\circ}$ and $(125 + 2 a)^{\circ}$ . If each one is the supplement of the other, then the value of a is:

45^{\circ}

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35^{\circ}

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25^{\circ}

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65^{\circ}

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Solution

The correct option is C $25^{\circ}$
Sum of supplementary angles is $180^{\circ}$
$∴ {(30 - a)}^{\circ} + {(125 + 2 a)}^{\circ} = 180^{\circ} \Rightarrow a + 155^{\circ} = 180^{\circ} \Rightarrow a = 180^{\circ} - 155^{\circ} = 25^{\circ}$

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Q. Two angles of a triangle measure 30° and 25° respectively. The measure of the third angle is
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(d) 125°

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Let N = $\frac{4^{5} + 4^{5} + 4^{5} + 4^{5}}{3^{5} + 3^{5} + 3^{5}} \cdot \frac{6^{5} + 6^{5} + 6^{5} + 6^{5} + 6^{5} + 6^{5}}{2^{5} + 2^{5}}$ then the value of ${log}_{2} N$ =

Q. Let

N = \frac{4^{5} + 4^{5} + 4^{5} + 4^{5}}{3^{5} + 3^{5} + 3^{5}} . \frac{6^{5} + 6^{5} + 6^{5} + 6^{5} + 6^{5} + 6^{5}}{2^{5} + 2^{5}}

, then the value of

{log}_{2} N =

Q. If

\frac{4^{5} + 4^{5} + 4^{5} + 4^{5}}{2^{5} + 2^{5}} \times \frac{6^{5} + 6^{5} + 6^{5} + 6^{5} + 6^{5} + 6^{5}}{3^{5} + 3^{5} + 3^{5}} = 8^{n}

, then n is:

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Two angles measure (30−a)∘ and (125+2a)∘ . If each one is the supplement of the other, then the value of a is:

The correct option is C 25∘Sum of supplementary angles is 180∘∴(30−a)∘+(125+2a)∘=180∘⇒a+155∘=180∘⇒a=180∘−155∘=25∘

Two angles measure $(30 - a)^{\circ}$ and $(125 + 2 a)^{\circ}$ . If each one is the supplement of the other, then the value of a is:

The correct option is C $25^{\circ}$
Sum of supplementary angles is $180^{\circ}$
$∴ {(30 - a)}^{\circ} + {(125 + 2 a)}^{\circ} = 180^{\circ} \Rightarrow a + 155^{\circ} = 180^{\circ} \Rightarrow a = 180^{\circ} - 155^{\circ} = 25^{\circ}$