A geometry problem by Luna Biswas

Geometry Level 3

Given any quadrilateral $ABCD$ $\angle DAB=150$ , $\angle DAC + \angle ABD=120$ , $\angle DBC - \angle ABD=60$

Find $\angle BDC$ .

The answer is 30.

1 solution

Luna Biswas
Mar 5, 2016

Denote <ABD= a and <BDC=b. By hypothesis <DAC= 120-a, <BAC= 30+a, <ADB= 30-a and <DBC= 60+a. By the law of sines on triangles ABC, BCD and ACD we get

BC/AC=sin(30+a)/sin(60+2a)=1/2cos(30+a)
DC/BC=sin(60+a)/sin b
AC/DC=sin(30-a+b)/sin(120-a)
Multiplying all these equalities, we have sin(30-a+b)=2cos(30+a)sin b
or 2cos(60+a)sin(30-b)=0

Thus <BDC=b=30

Could you provide a complete solution? Thanks!

Calvin Lin Staff - 5 years, 3 months ago

Yeah. Sure. Here it goes Denote <ABD= a and <BDC=b. By hypothesis <DAC= 120-a, <BAC= 30+a, <ADB= 30-a and <DBC= 60+a. By the law of sines on triangles ABC, BCD and ACD we get

BC/AC=sin(30+a)/sin(60+2a)=1/2cos(30+a) DC/BC=sin(60+a)/sin b AC/DC=sin(30-a+b)/sin(120-a) Multiplying all these equalities, we have sin(30-a+b)=2cos(30+a)sin b or 2cos(60+a)sin(30-b)=0

Thus <BDC=b=30

Luna Biswas - 5 years, 3 months ago

I forgot to give commas between the ratio of the sides. Sorry for the inconveniences

Luna Biswas - 5 years, 3 months ago

@Luna Biswas – Great thanks! This is wonderful :)

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Calvin Lin Staff - 5 years, 3 months ago

A geometry problem by Luna Biswas

The answer is 30.

1 solution

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