Question

a. Express $15\sin \theta - 8\cos \theta $ in the form $R\sin (\theta - \alpha )$, where \[R > 0\] and $0^\circ < \alpha < 90^\circ .$
Give the value of $\alpha $ correct to \[2\] decimal places.
b. Hence solve the equation $15\sin \theta - 8\cos \theta = 10$ for $0^\circ < \theta < 360^\circ .$

Answer 1

Hint: o solve this question, we will start with expressing the given expression $15\sin \theta - 8\cos \theta $ in the form $R\sin (\theta - \alpha )$, now we will construct the right-angled triangle, and then with the help of that figure we will get the value of $\alpha $. Now in the b part, to find the value of $\theta $, we will use the expression of a part, and on putting the value, we will get the required answer.

Complete step-by-step answer:
a. We have been given an expression $15\sin \theta - 8\cos \theta $ we need to express it in the form $R\sin (\theta - \alpha )$, where R>0 and $0^\circ < \alpha < 90^\circ .$ And then we need to give the value of $\alpha $ correct to 2 decimal places.
So, we are given the expression $ = 15\sin \theta - 8\cos \theta $
On multiplying and dividing whole expression by 17, we get
$ = 17(\dfrac{{15\sin \theta }}{{17}} - \dfrac{{8\cos \theta }}{{17}})$
$ = 17\sin (\theta - \alpha )$\[ \ldots .eq.\left( 1 \right)\]
So, we get the expression $15\sin \theta - 8\cos \theta $ in the form $R\sin (\theta - \alpha )$.
Now, we will draw a right-angled triangle, to get the value of $\alpha .$

$\cos \alpha = \dfrac{{15}}{{17}}$
$\sin \alpha = \dfrac{8}{{17}}$
$\tan \alpha = \dfrac{8}{{15}}$
$\alpha = ta{n^{ - 1}}\dfrac{8}{{15}}$
$ \Rightarrow \alpha = 28.07$
Thus, the value of $\alpha $ correct to 2 decimal places is 28.07.

b. Now, we need to solve the equation $15\sin \theta - 8\cos \theta = 10$ for $0^\circ < \theta < 360^\circ .$
So, the given equation is $15\sin \theta - 8\cos \theta = 10$
From \[eq.\left( 1 \right),\] we get
$\begin{gathered}
  17\sin (\theta - \alpha ) = 10 \\
  \sin (\theta - \alpha ) = \dfrac{{10}}{{17}} \\
\end{gathered} $
$\theta = \alpha + si{n^{ - 1}}\dfrac{{10}}{{17}}$
Now, on putting the value of $\alpha $from a part and the value of $si{n^{ - 1}}\dfrac{{10}}{{17}},$ we get
$\theta = 28.07 + 36.03$
$ \Rightarrow \theta = 64^\circ $, $0^\circ < \theta < 360^\circ .$
Thus, solution of $15\sin \theta - 8\cos \theta = 10$ for $0^\circ < \theta < 360^\circ ,$is $64^\circ .$

Note: Students should note that, in the solution, we have put the value of $ta{n^{ - 1}}\dfrac{8}{{15}}$ and $si{n^{ - 1}}\dfrac{{10}}{{17}},$ to get these inverse values, we took the help of a calculator. Though there is a long method to solve inverses without calculators, that you will learn in higher classes.