Par-Pen-Cooler

Geometry Level 4

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

B C = sin A × x BC=\sin\angle A×x


The answer is 8.

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4 solutions

Akshat Sharda
Jan 7, 2016

sin A = sin ( x + y ) = sin x cos y + cos x sin y = B D A B A D A C + A D A B C D A C = B D 8 + C D 8 = B C 8 sin A × 8 = B C \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 A D AD is a perpendicular dropped from A A on B C BC .

Akshat Sharda - 5 years, 5 months ago
Siddharth Singh
Nov 22, 2015

I created this question using the following equation:

A B A C = 2 R A D AB*AC=2R*AD ( R R is the circumradius of A B C \displaystyle \triangle ABC )

putting the values in the equation we get,

2 R = 8 2R=8

Now, using the sine rule

B C sin A = 2 R \frac{BC}{\sin\angle A}=2R

B C = 2 R sin A BC=2R*\sin\angle A

x = 8 x=8

Mohit Gupta
Nov 21, 2015

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

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. B C = A C S i n B A C = B C A C B y S i n L a w , B C S i n B A C = B C B C A C = 8. 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.

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