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

Hint: 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 ,
\[\begin{align}
  & \log \tan \left( \dfrac{B-C}{2} \right)=\log \left( \dfrac{2}{10}\cot \left( {{33}^{\circ }}{{20}^{'}} \right) \right) \\
 & =\log 2-\log 10+\log \cot \left( {{33}^{\circ }}{{20}^{'}} \right) \\
 & =.30103-1+10.18197=9.48300 \\
\end{align}\]
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 ,
\[\begin{align}
  & \cos {{16}^{\circ }}{{54}^{'}}{{50}^{''}}=\dfrac{237+158}{a}\sin \dfrac{{{66}^{\circ }}{{40}^{'}}}{2} \\
 & \Rightarrow a=395\sin {{33}^{\circ }}{{20}^{'}}\sec {{16}^{\circ }}{{54}^{'}}{{50}^{''}} \\
 & \Rightarrow a=\dfrac{79\times 10}{2}\sin {{33}^{\circ }}{{20}^{'}}\sec {{16}^{\circ }}{{54}^{'}}{{50}^{''}} \\
\end{align}\]
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)
\[\begin{align}
  & \Rightarrow \log a=2.89763-.30103+9.73998+10.01920 \\
 & \Rightarrow \log a=\text{23}\text{.35578} \\
\end{align}\]
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.