Find the Angle #3

Geometry Level 4

If,

C A D = B A C = 1 0 \angle CAD = \angle BAC = 10^\circ ,

D B A = 6 0 \angle DBA = 60^\circ , and

C B D = 8 0 \angle CBD = 80^\circ

Find D C A \angle DCA

This problem is a part of the set Find the Angle!


The answer is 20.

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2 solutions

Despicable Tamim
Apr 24, 2015

I used calculator when needed . I can't type the answer , instead I uploaded a photo of the ANSWER .

Let angle DAC=x ,,,,,

very lengthy process ......

Apurbo Sun - 6 years, 1 month ago

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level 3 problem man

Despicable Tamim - 6 years, 1 month ago

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level 4 dude

Syed Hamza Khalid - 4 years, 1 month ago

L e t x = D C A . S o C D O = 70 x . ( ) i n d i c a t e A p p l i c a t i o n o f S i n L a w . I n Δ D A B A O i s t h e a n g l e b i s e c t o r o f D A B . S o B O O D = B A A D = S i n 100 S i n 60 . . ( ) . . . . . . . . . . . . . ( A ) I n Δ D O C , O C = O D S i n ( 70 x ) S i n ( x ) . . ( ) . . . . . . . . ( B ) I n Δ C O B , O C = O B S i n 80 S i n 30 . . ( ) . . . . . . . . ( C ) F r o m ( B ) a n d ( C ) B O O D = S i n 30 S i n 80 S i n ( 70 x ) S i n ( x ) = S i n 100 S i n 60 . . b y ( A ) E x p a n d i n g S i n a n d s i m p l i f y i n g , S i n 70 C o t ( x ) C o s 70 = S i n 100 S i n 60 S i n 80 S i n 30 . x = T a n 1 ( S i n 70 S i n 100 S i n 80 S i n 60 S i n 30 + C o s 70 . ) D C A = 20. Let~x=\angle~DCA.~~So~\angle~CDO=70-x.\\ (*)~indicate~Application~of~Sin~Law.\\ ~~~~~\\ In~\Delta~DAB~AO~is~the~angle~bisector~of~\angle~DAB.\\ So~\dfrac{BO}{OD}=~\dfrac{BA}{AD}=\dfrac{Sin100}{Sin60}..(*).............(A)\\ In~\Delta~DOC,~~OC=OD*\dfrac{Sin(70-x)}{Sin(x)}..(*)........(B)\\ In~\Delta~COB,~~OC=OB*\dfrac{Sin80}{Sin30}..(*)........(C)\\ ~~~~\\ ~~~~~~~\\ From~(B)~and~(C)~~ \dfrac{BO}{OD}=\dfrac{Sin30}{Sin80}*\dfrac{Sin(70-x)}{Sin(x)}=\dfrac{Sin100}{Sin60}..by~~(A)\\ ~~~~~ \\ ~~~~~\\ Expanding~Sin~and~simplifying,\\ Sin70*Cot(x)-Cos70= \dfrac{Sin100}{Sin60}*\dfrac{Sin80}{Sin30}.\\ \therefore~~x~=~Tan^{-1} \left (\dfrac{Sin70}{\dfrac{Sin100*Sin80}{Sin60*Sin30}+Cos70}. \right )\\ \angle~DCA=\Large~~~~\color{#D61F06}{20}.

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