Question

If sin80°-
cos 70=
cosx°,then x/10=
​

Answer 1

Given:

$\sin 80^\circ-\cos 70^\circ=\cos x^\circ$.

To find:

The value of $\dfrac{x}{10}$.

Solution:

We have,

$\sin 80^\circ-\cos 70^\circ=\cos x^\circ$

$\sin 80^\circ-\cos (90^\circ-20^\circ)=\cos x^\circ$

$\sin 80^\circ-\sin (20^\circ)=\cos x^\circ$    $[\because \cos (90^\circ -\theta) = \sin \theta]$

$2\sin (\dfrac{80^\circ -20^\circ}{2})\cos (\dfrac{80^\circ+20^\circ}{2})=\cos x^\circ$      $[\because \sin A-\sin B=2\sin(\dfrac{A-B}{2})\cos (\dfrac{A+B}{2})]$

$2\sin (30^\circ)\cos (50^\circ)=\cos x^\circ$

$2\times \dfrac{1}{2}\times \cos (50^\circ)=\cos x^\circ$       $[\because \sin (30^\circ)=\dfrac{1}{2}]$

$\cos (50^\circ)=\cos x^\circ$

On comparting both sides, we get

$x=50$

Now,

$\dfrac{x}{10}=\dfrac{50}{10}=5$

Therefore, the required value is 5.

Answer 2

Given:

$sin80^\circ-cos70^\circ=cosx^\circ$

To find:

$\dfrac{x}{10}=?$

Solution:

Let us have a look at a few formula first:

1. $cos(90^\circ-\theta)=sin\theta$

2. $sinA-sinB=2cos(\dfrac{A+B}{2})sin(\dfrac{A-B}{2})$

Using first formula in Left Hand Side(LHS) of the given expression:

$cos70^\circ$ can be written as $cos(90^\circ-20^\circ)$

So, $cos70^\circ=sin20^\circ$

Therefore, the given expression becomes:

$sin80^\circ-sin20^\circ$

Now, using the second formula:

$sin80^\circ-sin20^\circ==2cos(\dfrac{80^\circ+20^\circ}{2})sin(\dfrac{80^\circ-20^\circ}{2})\\\Rightarrow 2cos50^\circ sin30^\circ$

We know that:

$sin30^\circ=\dfrac{1}{2}$

So, the expression becomes:

$2cos50^\circ \times \dfrac{1}{2}\\\Rightarrow cos50^\circ$

Comparing with Right Hand Side(RHS):

$cos50^\circ=cosx^\circ\\\Rightarrow x =50^\circ$

Dividing by 10:

$\dfrac{x}{10}=\dfrac{50^\circ}{10}\\\Rightarrow \dfrac{x}{10}=\bold{5^\circ}$