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mars1129 [50]
3 years ago
5

In this triangle, the product of sin B and tan C is and the product of sin Cand tan B is

Mathematics
1 answer:
Ipatiy [6.2K]3 years ago
7 0

Answer:

\frac{c}{a} and \frac{b}{a}

Step-by-step explanation:

sinB = \frac{opposite}{hypotenuse} = \frac{AC}{BC} = \frac{b}{a}

tanC = \frac{opposite}{adjacent} = \frac{AB}{AC} = \frac{c}{b}

Thus

sinB tanC = \frac{b}{a} × \frac{c}{b} ( cancel b on numerator/ denominator )

                 = \frac{c}{a}

---------------------------------------------------------------------------

sinC = \frac{opposite}{hypotenuse} = \frac{AB}{BC} = \frac{c}{a}

tanB = \frac{opposite}{adjacent} = \frac{AC}{AB} = \frac{b}{c}

Thus

sinC tanB = \frac{c}{a} × \frac{b}{c} ( cancel c on numerator/ denominator )

                 = \frac{b}{a}

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Given: AB ≅ BC , m∠MOC = 135° OM − angle bisector of ∠AOB Prove: ∠ABO ≅ ∠CBO
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Answer:

See explanation

Step-by-step explanation:

1. Angles AOM and MOC are supplementry angles. If m∠MOC = 135°, then

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3. Now

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3 years ago
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Answer:

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Step-by-step explanation:

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  234/3 = 78

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3 years ago
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Answer:

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