Question-216245

Question Number 216245 by Tawa11 last updated on 01/Feb/25

Commented by Tawa11 last updated on 01/Feb/25

I can use cosine rule .

But I want to see if there is

better approach .

Commented by Tawa11 last updated on 01/Feb/25

I got x + y = 50 .

But to proceed gives me maths error .

Commented by mr W last updated on 02/Feb/25

w e c a n f i n d x = 30 ° p u r e l y

g e o m e t r i c a l l y w i t h o u t u s i n g s i n e o r

c o s i n e r u l e s .

s h o w y o u r s o l u t i o n u s i n g c o s i n e r u l e .

Commented by Tawa11 last updated on 01/Feb/25

Commented by Tawa11 last updated on 01/Feb/25

I represent those equal length as 1 .

I leant that from you sir .

BD = \sqrt{2 - 2 \cos 40}

Then, I found BC .

BC = 1^{2} + {(\sqrt{2 - 2 \cos 40})}^{2} - 2 (1) (2 - 2 \cos 40) \cos 130

BC = 3 - 2 \cos 40 - 2 (2 - 2 \cos 40) \cos 130

So, \frac{\sin x}{1} = \frac{\sin 130}{3 - 2 \cos 40 - 2 (2 - 2 \cos 40) \cos 130}

So, I am getting x approximately 32 now .

Commented by mr W last updated on 02/Feb/25

w r o n g!

{BC}^{2} = 1^{2} + {(\sqrt{2 - 2 \cos 40})}^{2} - 2 (1) \sqrt{2 - 2 \cos 40} \cos 130

Commented by Tawa11 last updated on 02/Feb/25

Ohhh .

How I did not see that ?

Thank you sir, I really appreciate .

Answered by mr W last updated on 01/Feb/25

Commented by mr W last updated on 02/Feb/25

b l u e n u m b e r s i n d i c a t e t h e s t e p s

x = 70 - 40 = 30 °

Commented by Tawa11 last updated on 02/Feb/25

Thanks sir .

I appreciate .

Commented by ArshadS last updated on 02/Feb/25

S i r I {c o u d n}^{'} t u n d e r s t a n d s t e p 5 . H o w i s i t 40 ?

Commented by mr W last updated on 02/Feb/25

∠ A E D = 60 + 40 = 100 °

E A = E D

\Rightarrow ∠ E A D = ∠ E D A = \frac{180 - 100}{2} = 40 °

Commented by ArshadS last updated on 03/Feb/25

T han k s S ir! I understood now .

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