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.

$$\mathrm{I}\:\mathrm{can}\:\mathrm{use}\:\mathrm{cosine}\:\mathrm{rule}. \\ $$$$\mathrm{But}\:\mathrm{I}\:\mathrm{want}\:\mathrm{to}\:\mathrm{see}\:\mathrm{if}\:\mathrm{there}\:\mathrm{is} \\ $$$$\mathrm{better}\:\mathrm{approach}. \\ $$

Commented by Tawa11 last updated on 01/Feb/25

I got x + y = 50. But to proceed gives me maths error.

$$\mathrm{I}\:\mathrm{got}\:\:\:\mathrm{x}\:\:+\:\:\mathrm{y}\:\:=\:\:\mathrm{50}. \\ $$$$\mathrm{But}\:\mathrm{to}\:\mathrm{proceed}\:\mathrm{gives}\:\mathrm{me}\:\:\mathrm{maths}\:\mathrm{error}. \\ $$

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

we can find x=30° purely geometrically without using sine or cosine rules. show your solution using cosine rule.

$${we}\:{can}\:{find}\:{x}=\mathrm{30}°\:{purely} \\ $$$${geometrically}\:{without}\:{using}\:{sine}\:{or}\: \\ $$$${cosine}\:{rules}. \\ $$$${show}\:{your}\:{solution}\:{using}\:{cosine}\:{rule}. \\ $$

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 = (√(2 − 2cos40)) Then, I found BC. BC = 1^2 + ((√(2 − 2cos40)))^2 − 2(1)(2 − 2cos40)cos130 BC = 3 − 2cos40 − 2(2 − 2cos40)cos130 So, ((sin x)/1) = ((sin130)/(3 − 2cos40 − 2(2 − 2cos40)cos130)) So, I am getting x approximately 32 now.

$$\mathrm{I}\:\mathrm{represent}\:\mathrm{those}\:\mathrm{equal}\:\mathrm{length}\:\mathrm{as}\:\mathrm{1}. \\ $$$$\mathrm{I}\:\mathrm{leant}\:\mathrm{that}\:\mathrm{from}\:\mathrm{you}\:\mathrm{sir}. \\ $$$$\:\:\:\:\:\:\:\mathrm{BD}\:\:=\:\:\sqrt{\mathrm{2}\:\:−\:\:\mathrm{2cos40}} \\ $$$$\mathrm{Then},\:\mathrm{I}\:\mathrm{found}\:\mathrm{BC}. \\ $$$$\:\:\:\:\:\:\:\mathrm{BC}\:\:=\:\:\mathrm{1}^{\mathrm{2}} \:\:+\:\:\left(\sqrt{\mathrm{2}\:\:−\:\:\mathrm{2cos40}}\right)^{\mathrm{2}} \:\:−\:\:\mathrm{2}\left(\mathrm{1}\right)\left(\mathrm{2}\:\:−\:\:\mathrm{2cos40}\right)\mathrm{cos130} \\ $$$$\:\:\:\:\:\:\:\mathrm{BC}\:\:=\:\:\mathrm{3}\:\:−\:\:\mathrm{2cos40}\:\:−\:\:\mathrm{2}\left(\mathrm{2}\:\:−\:\:\mathrm{2cos40}\right)\mathrm{cos130} \\ $$$$\mathrm{So},\:\:\:\:\:\:\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{1}}\:\:=\:\:\frac{\mathrm{sin130}}{\mathrm{3}\:\:−\:\:\mathrm{2cos40}\:\:−\:\:\mathrm{2}\left(\mathrm{2}\:\:−\:\:\mathrm{2cos40}\right)\mathrm{cos130}} \\ $$$$\mathrm{So},\:\mathrm{I}\:\mathrm{am}\:\mathrm{getting}\:\:\:\mathrm{x}\:\:\mathrm{approximately}\:\:\:\mathrm{32}\:\:\:\mathrm{now}. \\ $$

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

wrong! BC^2 = 1^2 + ((√(2 − 2cos40)))^2 − 2(1)(√(2 − 2cos40)) cos130

$${wrong}! \\ $$$$\:\:\:\:\:\:\:\mathrm{BC}^{\mathrm{2}} \:\:=\:\:\mathrm{1}^{\mathrm{2}} \:\:+\:\:\left(\sqrt{\mathrm{2}\:\:−\:\:\mathrm{2cos40}}\right)^{\mathrm{2}} \:\:−\:\:\mathrm{2}\left(\mathrm{1}\right)\sqrt{\mathrm{2}\:\:−\:\:\mathrm{2cos40}}\:\mathrm{cos130} \\ $$

Commented by Tawa11 last updated on 02/Feb/25

Ohhh. How I did not see that? Thank you sir, I really appreciate.

$$\mathrm{Ohhh}. \\ $$$$\mathrm{How}\:\mathrm{I}\:\mathrm{did}\:\mathrm{not}\:\mathrm{see}\:\mathrm{that}? \\ $$$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir},\:\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$

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

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

blue numbers indicate the steps x=70−40=30°

$${blue}\:{numbers}\:{indicate}\:{the}\:{steps} \\ $$$${x}=\mathrm{70}−\mathrm{40}=\mathrm{30}° \\ $$

Commented by Tawa11 last updated on 02/Feb/25

$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{appreciate}. \\ $$

Commented by ArshadS last updated on 02/Feb/25

Sir I coudn′t understand step 5. How is it 40 ?

$${Sir}\:{I}\:\:{coudn}'{t}\:{understand}\:{step}\:\:\mathrm{5}.\:{How}\:{is}\:{it}\:\:\mathrm{40}\:? \\ $$

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

∠AED=60+40=100° EA=ED ⇒∠EAD=∠EDA=((180−100)/2)=40°

$$\angle{AED}=\mathrm{60}+\mathrm{40}=\mathrm{100}° \\ $$$${EA}={ED}\: \\ $$$$\Rightarrow\angle{EAD}=\angle{EDA}=\frac{\mathrm{180}−\mathrm{100}}{\mathrm{2}}=\mathrm{40}° \\ $$

Commented by ArshadS last updated on 03/Feb/25

$$\mathbb{T}\boldsymbol{\mathrm{han}}\Bbbk\boldsymbol{\mathrm{s}}\:\mathbb{S}\boldsymbol{\mathrm{ir}}!\:\mathrm{I}\:\mathrm{understood}\:\mathrm{now}. \\ $$

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