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3 years ago

Given the triangle below, what is m < B, rounded to the nearest tenth

Mathematics

2 answers:

aalyn [17]3 years ago

4 0

Answer:

Therefore m∠ B is 42.1°

Step-by-step explanation:

Given:

In Δ ABC

AB = 15 = c

AC = 18 = b

∠ C = 34°

To Find:

∠ B = ?

Solution:

In Δ ABC We have Sine Rule as

$\frac{a}{\sin A}= \frac{b}{\sin B}= \frac{c}{\sin C}$

Substituting the given values we get

$\frac{b}{\sin B}= \frac{c}{\sin C}$

$\frac{18}{\sin B}= \frac{15}{\sin 34}\\\\\sin B=0.671\\\\\therefore B=\sin^{-1}(0.671)=42.14\°$

Therefore m∠ B is 42.1°

solmaris [256]3 years ago

4 0

Answer:42.1

Step-by-step explanation:

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Vera_Pavlovna [14]

Split up the integration interval into 4 subintervals:

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We approximate the (signed) area under the curve over each subinterval by

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$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4M_i\approx\boxed{2.981137}$

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We first interpolate the integrand over each subinterval by a quadratic polynomial $p_i(x)$ , where

$p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}$

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$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx$

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$S_i=\displaystyle\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx=\frac{r_i-\ell_i}6(f(\ell_i)+4f(m_i)+f(r_i))$

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$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4S_i\approx\boxed{3.000117}$

Compare these to the actual value of the integral, 3. I've included plots of the approximations below.

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