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2 years ago

Plz help!! Consider the given right triangle.

Mathematics

1 answer:

aalyn [17]2 years ago

4 0

Answer:

b = 8

c = 16

Step-by-step explanation:

From the given right triangle,

a = $8\sqrt{3}$ , m(∠B) = 30°

By applying sine rule in the given triangle,

cos(B) = $\frac{\text{Adjacent side}}{\text{Hypotenuse}}$

cos(B) = $\frac{BC}{AB}$

cos(30°) = $\frac{8\sqrt{3}}{c}$

$\frac{\sqrt{3} }{2}= \frac{8\sqrt{3} }{c}$

c = 16

Further we apply tangent rule,

tan(B) = $\frac{\text{Opposite side}}{\text{Adjacent side}}$

tan(30°) = $\frac{b}{a}$

$\frac{1}{\sqrt{3}}=\frac{b}{8\sqrt{3} }$

b = 8

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2 years ago

If a=2b^3 and b= -1/2c ^-2,express a in terms of c.

ElenaW [278]

Answer:

Part 1) $a=-\frac{1}{4c^6}$

Part 2) $a=-\frac{1}{4c^{-6}}$

Step-by-step explanation:

Part 1) we have

$a=2b^{3}$ ----> equation A

$b=-\frac{1}{2}c^{-2}$ ----> equation B

substitute equation B in equation A

$a=2(-\frac{1}{2}c^{-2})^{3}$

Applying property of exponents

$(x^{m})^{n}=x^{m*n}$

$x^{-m} =\frac{1}{x^{m}}$

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therefore

$a=-\frac{1}{4c^6}$

Part 2) we have

$a=2b^{3}$ ----> equation A

$b=-\frac{1}{2c^{-2}}=-\frac{c^{2}}{2}$ ----> equation B

substitute equation B in equation A

$a=2(-\frac{c^{2}}{2})^{3}$

Applying property of exponents

$(x^{m})^{n}=x^{m*n}$

$x^{-m} =\frac{1}{x^{m}}$

$a=2(-\frac{c^{2}}{2})^{3}=2(-\frac{c^{6}}{8})$

simplify

$a=-\frac{c^{6}}{4}$

therefore

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