Question

in a right-angle triangle \[{\text{ABC}}\] where right angle at \[{\text{C}}\], \[{\text{BC + CA = 23}}\;{\text{cm}}\] and \[{\text{BC}} - {\text{CA = 7}}\;{\text{cm}}\]. Then find $\sin {\text{A}}$ and $\tan {\text{B}}$ .

Answer 1

Hint: Find the value of \[{\text{BC}}\] or \[{\text{CA}}\] by cancellation law apply on those equations, draw a right-angle triangle and also find the value of \[{\text{AB}}\] by Pythagoras theorem then use the formulas of \[\sin \;{\text{A}}\;{\text{and tan}}\;{\text{B}}\].

Complete step-by-step solution:
Given: The provided first equation is \[{\text{BC + CA = 23}}\;{\text{cm}}\] and \[{\text{BC - CA}} = 7cm\]
First, we will find the values of BC and CA by the two given equations.
Now, subtract \[{\text{CA}}\] from the both sides in equation \[{\text{BC + CA = 23}}\;{\text{cm}}\]
\[
  {\text{BC + CA}}\; - {\text{CA = 23}}\; - {\text{CA}} \\
  {\text{BC = 23}}\; - {\text{CA}} \\
\]
Put the value \[{\text{BC}}\]in the second equation\[{\text{BC}}\; - {\text{CA = 7}}\;{\text{cm}}\]as:
\[23 - {\text{CA}}\; - {\text{CA = 7}}\;{\text{cm}}\]
Solve further,
\[
   - 2{\text{CA}}\;{\text{ = 7}}\; - 23 \\
  {\text{CA}} = 8\;{\text{cm}} \\
\]
For the value of \[{\text{BC}}\]put the value of \[{\text{CA}}\]in \[{\text{BC = 23}}\; - {\text{CA}}\],
\[
  {\text{BC = 23}}\; - 8 \\
  {\text{BC = }}15\;{\text{cm}} \\
\]
Now draw a right angle triangle whose sides are\[{\text{CA}} = 8\;{\text{cm}}\]and \[{\text{BC = }}15\;{\text{cm}}\]and then we have to find the value of \[{\text{AB}}\] by rule of Pythagoras theorem,
Pythagoras rule: \[{\left( {{\text{base}}} \right)^2} + {\left( {{\text{height}}} \right)^2} = {\left( {{\text{hypotenuse}}} \right)^2}\]

base\[{\text{BC = }}15\;{\text{cm}}\],height\[{\text{CA}} = 8\;{\text{cm}}\]now use Pythagoras theorem,
\[
  {\left( {{\text{base}}} \right)^2} + {\left( {{\text{height}}} \right)^2} = {\left( {{\text{hypotenuse}}} \right)^2} \\
  {\left( {{\text{CB}}} \right)^2} + {\left( {{\text{AC}}} \right)^2} = {\left( {{\text{AB}}} \right)^2} \\
  {\left( {15} \right)^2} + {\left( 8 \right)^2} = {\left( {{\text{AB}}} \right)^2} \\
  225 + 64 = {\left( {{\text{AB}}} \right)^2} \\
\]
Furthermore,
\[
  289 = {\left( {{\text{AB}}} \right)^2} \\
  \left( {{\text{AB}}} \right) = \sqrt {289} \\
  \left( {{\text{AB}}} \right) = 17 \\
\]
Now we have base, height and hypotenuse so we solve for the values of \[\sin \;{\text{A}}\;{\text{and tan}}\;{\text{B}}\].
First solve for $\sin {\text{A}}$ .
\[
  \sin \;{\text{A}} = \dfrac{{{\text{base}}}}{{{\text{hypothesis}}}} \\
  \sin \;{\text{A}} = \dfrac{{{\text{BC}}}}{{{\text{AB}}}} \\
  \sin \;{\text{A}} = \dfrac{{15\;cm}}{{17\;cm}} \\
\]
Now solve for \[{\text{tan}}\;{\text{B}}\];
\[
  \tan \;{\text{B}} = \dfrac{{{\text{height}}}}{{{\text{base}}}} \\
  \tan \;{\text{B}} = \dfrac{{{\text{AC}}}}{{{\text{BC}}}} \\
  \tan \;{\text{B}} = \dfrac{{8\;cm}}{{15\;cm}} \\
\]
Thus, the values of \[\sin \;{\text{A}} = \dfrac{{15\;cm}}{{17\;cm}}\] and \[\tan \;{\text{B}} = \dfrac{{8\;cm}}{{15\;cm}}\].
Note: First find all the three sides of a right angle triangle and then find $\sin {\text{A}}$ and $\tan {\text{B}}$ by using the properties of the triangle .