Question

Two sides of a triangle are 237 and 158 feet and the contained angle is ${{66}^{\circ }}{{40}^{'}}$; find the base and the other angles, having given $\log 2=.3013,\log 79=1.89763$, $\log 1.36383=23.35578,\log \cot {{33}^{\circ }}{{20}^{'}}=10.18197,\log \sin {{33}^{\circ }}{{20}^{'}}=9.73998$,$\log \tan {{16}^{\circ }}{{54}^{'}}=9.48262$,$\log \tan {{16}^{\circ }}{{55}^{'}}=9.48308,\log \sec {{16}^{\circ }}{{54}^{'}}=10.01917,\log \sec {{16}^{\circ }}{{55}^{'}}=10.01921$ $$$$

Answer 1

We use the law of tangent $\tan \left( \dfrac{B-C}{2} \right)=\dfrac{b-c}{b+c}\cot \dfrac{A}{2}$ where we assign $b=237$feet and $c=158$feet,$A={{66}^{\circ }}{{40}^{'}}$ and then take logarithm to get the value of $\dfrac{B-C}{2}$. We then find $\dfrac{B+C}{2}=\dfrac{180-A}{2}$. We solve for $B,C$ and obtain them. We use the formula $\cos \dfrac{B-C}{2}=\dfrac{b+c}{a}\sin \dfrac{A}{2}$ and take logarithm to get $a$. We assume infinitesimal change in the functions $\log \tan x,\log \sec x$ as constant. $$$$

Complete step by step answer:
We know from the law of tangent that that if there are the measurement of three angles are denoted as $A,B,C$ of triangle ABC and the length their opposite sides are denoted as $a,b,c$ respectively , then we write the relation among them as
$\tan \left( \dfrac{B-C}{2} \right)=\dfrac{b-c}{b+c}\cot \dfrac{A}{2}$
Let us assign $b=237$feet and $c=158$. The angle contained between the sides whose lengths are $b,c$ is $A={{66}^{\circ }}{{40}^{'}}$ as given in the question . We put these values and proceed
$\tan \left( \dfrac{B-C}{2} \right)=\dfrac{237-158}{237+158}\cot \left( \dfrac{{{66}^{\circ }}{{40}^{'}}}{2} \right)=\dfrac{1}{5}\cot \left( {{33}^{\circ }}{{20}^{'}} \right)=\dfrac{2}{10}\cot \left( {{33}^{\circ }}{{20}^{'}} \right)$
We take logarithm both side and get ,

& \log \tan \left( \dfrac{B-C}{2} \right)=\log \left( \dfrac{2}{10}\cot \left( {{33}^{\circ }}{{20}^{'}} \right) \right) \\\ & =\log 2-\log 10+\log \cot \left( {{33}^{\circ }}{{20}^{'}} \right) \\\ & =.30103-1+10.18197=9.48300 \\\ \end{aligned}$$ We are given the data in the question $\log \tan {{16}^{\circ }}{{54}^{'}}=9.48262,\log \tan {{16}^{\circ }}{{55}^{'}}=9.48308$. We approximate the change in $f\left( x \right)=\log \tan x$ to be constant for very small value like $dx={{1}^{'}}$. We can see the change as $dx={{1}^{'}}={{60}^{''}}$ leads to change in value of $\log \tan x$ as $\log \tan {{16}^{\circ }}{{55}^{'}}-\log \tan {{16}^{\circ }}{{54}^{'}}=9.48308-9.48262=0.00046$. The change in $\log \tan x$ with respect to change in $x$ from $\log \tan {{16}^{\circ }}{{54}^{'}}=9.48262$ to $\log \tan \left( \dfrac{B-C}{2} \right)=9.483$ is $9.483-9.48262=0.00038$. The equivalent change in $x$ for change of $0.00038$ is $\dfrac{0.00038}{0.00046}\times {{60}^{''}}={{50}^{''}}$. So we have $$\dfrac{B-C}{2}={{16}^{\circ }}{{54}^{'}}+{{50}^{''}}={{16}^{\circ }}{{54}^{'}}{{50}^{''}}.....(1)$$ We know that $A+B+C={{180}^{\circ }}$. So we put the required value and obtain , $$\dfrac{B+C}{2}=\dfrac{180-A}{2}={{56}^{\circ }}{{40}^{'}}....(2)$$ We solve the linear pair of equation (1) and (2) and the measurement of the angles $B={{73}^{\circ }}{{34}^{'}}{{50}^{''}},C={{39}^{\circ }}{{45}^{'}}{{10}^{''}}$ We know dor the law of sine and cosine are interrelated as $$\cos \dfrac{B-C}{2}=\dfrac{b+c}{a}\sin \dfrac{A}{2}$$ We put the given values , the obtained value $\dfrac{B-C}{2}$ and get , $$\begin{aligned} & \cos {{16}^{\circ }}{{54}^{'}}{{50}^{''}}=\dfrac{237+158}{a}\sin \dfrac{{{66}^{\circ }}{{40}^{'}}}{2} \\\ & \Rightarrow a=395\sin {{33}^{\circ }}{{20}^{'}}\sec {{16}^{\circ }}{{54}^{'}}{{50}^{''}} \\\ & \Rightarrow a=\dfrac{79\times 10}{2}\sin {{33}^{\circ }}{{20}^{'}}\sec {{16}^{\circ }}{{54}^{'}}{{50}^{''}} \\\ \end{aligned}$$ We take logarithm both side and get , $$\Rightarrow \log a=\log 79+\log 10-\log 2+\log \sin {{33}^{\circ }}{{20}^{'}}+\log \sec {{16}^{\circ }}{{54}^{'}}{{50}^{''}}...(3)$$ We again denote $f\left( x \right)=\log \sec x$. We can assume that for every small change in $x$ the change in $f\left( x \right)$ will be constant. So we have data from the question $\log \sec {{16}^{\circ }}{{54}^{'}}=10.01917,\log \sec {{16}^{\circ }}{{55}^{'}}=10.01921$. So the change in $x={{1}^{'}}={{60}^{''}}$ is equivalent to change in $f\left( x \right)=\log \sec x$ as $\log \sec {{16}^{\circ }}{{55}^{'}}-\log \sec {{16}^{\circ }}{{54}^{'}}=10.01921-10.01917=0.00004$. So the change $x={{50}^{''}}$ will be $\dfrac{{{50}^{''}}}{{{60}^{''}}}\times .00004=0.00003$. So we have $$\log \sec {{16}^{\circ }}{{54}^{'}}{{50}^{''}}=10.0197+0.00003=10.01920$$. So now we proceed in equation(3) $$\begin{aligned} & \Rightarrow \log a=2.89763-.30103+9.73998+10.01920 \\\ & \Rightarrow \log a=\text{23}\text{.35578} \\\ \end{aligned}$$ So we got $a=1.3683$feet. $$$$ **Note:** We assumed the changes in functions $\log \tan x,\log \sec x$ as constant because the changes in $x$ were infinitesimally small. When the change is infinitesimally small we can find the rate of change of functions using derivatives.