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In an triangle the least angle is $45^\circ$ and the tangents of the angles are in $A.P.$If its area be $27$ sq.cm.Find the lengths of its sides.


Let $A$ be the smallest angle.$\angle A=45^\circ$.$\tan A,\tan B,\tan C$ are in arithmatic progression.

$2\tan B=\tan A+\tan C$

$2\frac{\sin B}{\cos B}=\frac{\sin B}{\cos A\cos C}$

$\cos A\cos C=\frac{\cos B}{2}$

$\frac{1}{\sqrt2}\cos C=\frac{\cos B}{2}$

$\frac{\cos B}{\cos C}=\sqrt2$

I am stuck here.Please help.

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  • $\begingroup$ @Shailesh he used $A+C=\pi -B$ $\endgroup$ – Nikunj Apr 23 '16 at 10:16
  • $\begingroup$ @Nikunj Of course. Thanks $\endgroup$ – Shailesh Apr 23 '16 at 11:57
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almagest has provided a good answer. This answer uses your idea.

Using that $B+C=180^\circ-45^\circ=135^\circ$, you can have $$\frac{\cos B}{\cos C}=\sqrt 2\quad\Rightarrow\quad \cos B=\sqrt 2\ \cos(135^\circ-B)=-\cos B+\sin B,$$ i.e.$$2\cos B=\sin B$$ Squaring the both sides gives $$4(1-\sin^2B)=\sin^2B\quad\Rightarrow\quad \sin B=\frac{2}{\sqrt 5},\quad \cos B=\frac{1}{\sqrt 5}\tag1$$ because $B\le 90^\circ$, and so we have $$\sin C=\sin(135^\circ-B)=\frac{3}{\sqrt{10}}\tag2$$

By the way, $$27=\frac 12bc\sin A\quad\Rightarrow\quad bc=54\sqrt 2$$ and so using the law of sines with $(1)(2)$ gives $$2R\sin B\cdot 2R\sin C=54\sqrt 2\quad\Rightarrow\quad R=\frac{3\sqrt 5}{\sqrt 2}$$where $R$ is the circumradius.

Thus, $$a=2R\sin A=3\sqrt 5,\quad b=2R\sin B=6\sqrt 2,\quad c=2R\sin C=9.$$

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Easier to use the $\tan(X+Y)$ formula.

You have $C=135^o-B$, so $\tan C=\frac{\tan135^o-\tan B}{1+\tan135^o\tan B}=\frac{1+t}{t-1}$, where $t=\tan B$. But since the tangents are in AP we have $\tan C=2t-1$, so we get the equation $2t^2-4t=0$. We cannot have $t=0$, so $t=2$. That gives us the angles. We have $\sin B=\frac{2}{\sqrt5}$ and $\tan C=3$, so $\sin C=\frac{3}{\sqrt{10}}$.

We are also given the area is 27. Suppose $a=k$. Then by the sine rule we have $b=\frac{2\sqrt2}{\sqrt5}k$ and $c=\frac{3\sqrt2}{\sqrt{10}}k=\frac{3}{\sqrt5}k$. Hence the area $27=\frac{1}{2}ab\sin C=\frac{\sqrt2}{\sqrt5}k^2\frac{3}{\sqrt{10}}=\frac{3}{5}k^2$. So $k=3\sqrt5$, giving $a=3\sqrt5,b=6\sqrt2,c=9$

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Another approach knowing that $$\tan A = 1$$ can be as follows: $$2\tan B = \tan A + \tan C$$ Dividing LHS and RHS by non-zero (1 - tan C), we get $$\frac{2\tan B}{1-\tan C} = \frac{1 +\tan C}{1 - \tan C} $$
$$<=>\frac{2\tan B}{1-\tan C} = \frac{\tan A +\tan C}{1 - \tan A \tan C}$$ $$<=>\frac{2\tan B}{1-\tan C} = \tan (A + C)$$ $$<=>\frac{2\tan B}{1-\tan C} = \tan (\pi - B)$$ $$<=>\frac{2\tan B}{1-\tan C} = -\tan B$$ Cancelling the common factors, we get $$\frac{2}{1-\tan C} = -1$$ Solving for tan C, we get $$\tan C = 3$$ Hence, $$\tan B = \frac{\tan A + \tan C}{2} = \frac{1 + 3}{2} = 2$$ Thus, knowing the value of tan B and tan C, we can calculate sin B and sin C. $$\sin B = \frac{2}{\sqrt 5} and \sin C = \frac{3}{\sqrt 10}$$ Also, we know by sine formula, $$b = 2R \sin B$$ and $$c = 2R \sin c$$ Rest is using the area Formula, $$27 = \frac{1}{2}bc \sin A$$ $$<=>27 = \frac{1}{2}(2R \sin B) .(2R \sin C) \sin A$$ $$<=>27 = \frac{1}{2}.4R^2 \frac{2}{\sqrt 5}. \frac{3}{\sqrt 10}.\frac{1}{\sqrt 2}$$ Solving for R, we get $$R = \frac {3\sqrt 5}{\sqrt 2}$$ Finally, $$ a = 2R \sin A = 2.\frac{3\sqrt 5}{\sqrt 2}.\frac{1}{\sqrt 2} = 3\sqrt 5 $$ $$ b = 2R \sin B = 2.\frac{3\sqrt 5}{\sqrt 2}.\frac{2}{\sqrt 5} = 6\sqrt 2 $$ $$ c = 2R \sin C = 2.\frac{3\sqrt 5}{\sqrt 2}.\frac{3}{\sqrt 10} = 9 $$

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$\sin A=\dfrac1{\sqrt2}$

As $B+C=180^\circ-A=(180-45)^\circ=\cdots,$

$$\cos B=\sqrt2\cos C=\sqrt2\cos(135^\circ-B)=\cdots$$

$$\iff\dfrac{\sin B}2=\dfrac{\cos B}1=\pm\dfrac1{\sqrt{2^2+1^2}}$$

As $\sin B>0,\sin B=+\dfrac2{\sqrt5},$

$\cos B=\dfrac1{\sqrt5}\implies\cos C=\dfrac{\cos B}{\sqrt2}=\dfrac1{\sqrt{10}}$

Again as $\sin C>0,\sin C=+\sqrt{1-\cos^2C}=\dfrac3{\sqrt{10}}$

Now use $\triangle=\dfrac{abc}{4R}=2R^2\sin A\sin B\sin C$ and sine rule

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