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Andrew [12]
4 years ago
8

In the diagram below AQRS is an equilateral triangle and RT

Mathematics
1 answer:
Minchanka [31]4 years ago
5 0

Answer:

The answer is D.

Step-by-step explanation:

AQRT is a 30-60-90 triangle.

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Verify tan(α+β) = [tan(α)+tan(β)] / [1 + tan(α)tan(β)]
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Answer:

Proved

Step-by-step explanation:

To Prove: tan(\alpha+\beta) =\dfrac{tan(\alpha)+tan(\beta)}{1 + tan(\alpha)tan(\beta)}

Proof:

Now: tan \theta =\dfrac{sin\theta }{cos \theta}

Therefore:

tan (\alpha+\beta)=\dfrac{ sin (\alpha+\beta)}{cos (\alpha+\beta) }

Applying these angle sum formula

sin (\alpha+\beta)=sin \alpha cos \alpha + sin \beta cos \beta\\cos (\alpha+\beta)=cos \alpha cos \beta - sin \alpha sin \beta

tan (\alpha+\beta)=\dfrac{ sin \alpha cos \alpha + sin \beta cos \beta}{cos \alpha cos \beta - sin \alpha sin \beta }

Divide all through by cos \alpha cos \beta

tan (\alpha+\beta)=\dfrac{ (sin \alpha cos \alpha)/(cos \alpha cos \beta) + (sin \beta cos \beta)/(cos \alpha cos \beta)}{(cos \alpha cos \beta)/(cos \alpha cos \beta) - (sin \alpha sin \beta)/(cos \alpha cos \beta) }\\\\tan (\alpha+\beta)=\dfrac{\frac{sin \alpha}{cos \alpha}+\frac{sin \beta}{cos \beta} }{1-tan \alpha tan \beta} \\$Therefore:\\\\tan (\alpha+\beta)=\dfrac{tan \alpha+tan \beta}{1-tan \alpha tan \beta}

=sin \alpha cos \beta + cos \alpha sin \beta/cos \alpha cos \beta/cos \alpha cos \beta- sin \alpha sin \beta/cos \alpha cos \beta  

=sin \alpha/cos \alpha + sin \beta/cos \beta/1-tan \alpha tan \beta  

=tan A + tan B/1-tan A tan B

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4 years ago
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I hope this helps!
~kaikers

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Substitute 8 for x
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