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3 years ago

Suppose an isosceles triangle ABC has A = 45° and b = c = 4. What is the length of a2? A.22.63

Mathematics

1 answer:

weeeeeb [17]3 years ago

3 0

Since b = c, the angles B and C also equal

so you get
180 - 45 = 135
135/2 = =67.5 for each angle

now we know 2 sides and all 3 angles, use the law of sines to find a

b/sin B = a/sin A

4/(sin 67.5) = a/(sin 45)

a = 4(sin 45)/(sin 67.5)

a^2 = (4(sin 45)/(sin 67.5))^2

a^2 = 9.37

Answer is C

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3 years ago

Read 2 more answers

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ElenaW [278]

Answer with Step-by-step explanation:

We are given that

$f(x)=\left\{\begin{matrix}\dfrac{x^2-2x}{x^2-4}&,if\ \ x\neq 2 \\ 1&,if\ \ x=2\end{matrix}\right.$

We have to explain that why the function is discontinuous at x=2

We know that if function is continuous at x=a then LHL=RHL=f(a).

$f(x)=\frac{x(x-2)}{(x+2)(x-2)}=\frac{x}{x+2}$

LHL=Left hand limit when x <2

Substitute x=2-h

where h is small positive value >0

$\lim_{h\rightarrow 0}f(x)=\lim_{h\rightarrow 0}\frac{2-h}{2-h+2}$

$\lim_{h\rightarrow 0}\frac{2-h}{4-h}=\frac{2}{4}=\frac{1}{2}$

Right hand limit =RHL when x> 2

Substitute

x=2+h

$\lim_{h\rightarrow 0}f(x)=\lim_{h\rightarrow 0}\frac{2+h}{2+h+2}=\lim_{h\rightarrow 0}\frac{2+h}{4+h}$

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LHL=RHL= $\frac{1}{2}$

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