Par-Pen-Cooler

Geometry Level 4

In a $\displaystyle\triangle ABC$ , $AB×AC=48$ and $AD$ is the perpendicular dropped from $A$ on $BC$ ,such that $AD=6$ .What should replace $x$ in the below equation.

$BC=\sin\angle A×x$

The answer is 8.

4 solutions

Akshat Sharda
Jan 7, 2016

$\begin{aligned}\sin \angle A = \sin (x+y) & = \sin x \cos y + \cos x \sin y \\ & = \frac{BD}{AB} \cdot \frac{AD}{AC} + \frac{AD}{AB} \cdot \frac{CD}{AC} \\ & = \frac{BD}{8} + \frac{CD}{8} \\ & = \frac{BC}{8} \\ \Rightarrow \sin \angle A \times 8 & = BC \end{aligned}$

Moderator note:

Good approach.

You seem to have made an implicit assumption that D is on the line segment BC. Must this be true? If not, how would the proof change?

Note: This is one reason why in geometry proofs, we make use of directed angles and directed line segments, to avoid such implicit assumptions.

It is given in the question that $AD$ is a perpendicular dropped from $A$ on $BC$ .

Akshat Sharda - 5 years, 5 months ago

Siddharth Singh
Nov 22, 2015

I created this question using the following equation:

$AB*AC=2R*AD$ ( $R$ is the circumradius of $\displaystyle \triangle ABC$ )

putting the values in the equation we get,

$2R=8$

Now, using the sine rule

$\frac{BC}{\sin\angle A}=2R$

$BC=2R*\sin\angle A$

$x=8$

Mohit Gupta
Nov 21, 2015

This problem could be solved using the formulae as A= $\frac{1}{2}$ x B x H and A=1/2x pro. Of opp.sides x sin(a)

Niranjan Khanderia
Nov 22, 2015

Since altitude is AD=6, AB, or AC minimum can be only 6. So let AB=6. So B and D coincides. AC=48/6=8. Angle B= angle D=90. $BC=AC~~~SinBAC=\dfrac{BC}{AC}\\ ~~By ~Sin ~Law, ~\dfrac{BC}{SinBAC}=\dfrac{BC}{\frac{BC}{AC}}=\Large~~~\color{#D61F06}{8}.$

Moderator note:

This solution makes an assumption and deals with a specific case, instead of the general case.

Par-Pen-Cooler

The answer is 8.

4 solutions

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