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

Commented by Tawa11 last updated on 01/Feb/25

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

Commented by Tawa11 last updated on 01/Feb/25

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

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

$$wecanfindx=30°purely$$$$geometricallywithoutusingsineor$$$$cosinerules.$$$$showyoursolutionusingcosinerule.$$

Commented by Tawa11 last updated on 01/Feb/25

Commented by Tawa11 last updated on 01/Feb/25

$$\mathrm{I}\mathrm{represent}\mathrm{those}\mathrm{equal}\mathrm{length}\mathrm{as}1.$$$$\mathrm{I}\mathrm{leant}\mathrm{that}\mathrm{from}\mathrm{you}\mathrm{sir}.$$$$\mathrm{BD}=\sqrt{2-2\cos 40}$$$$\mathrm{Then},\mathrm{I}\mathrm{found}\mathrm{BC}.$$$$\mathrm{BC}=1^2+{\left(\sqrt{2-2\cos 40}\right)}^2-2\left(1\right)(2-2\cos 40)\cos 130$$$$\mathrm{BC}=3-2\cos 40-2(2-2\cos 40)\cos 130$$$$\mathrm{So},\frac{\sin \mathrm{x}}{1}=\frac{\sin 130}{3-2\cos 40-2(2-2\cos 40)\cos 130}$$$$\mathrm{So},\mathrm{I}\mathrm{am}\mathrm{getting}\mathrm{x}\mathrm{approximately}32\mathrm{now}.$$

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

$$wrong!$$$${\mathrm{BC}}^2=1^2+{\left(\sqrt{2-2\cos 40}\right)}^2-2\left(1\right)\sqrt{2-2\cos 40}\cos 130$$

Commented by Tawa11 last updated on 02/Feb/25

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

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

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

$$bluenumbersindicatethesteps$$$$x=70-40=30°$$

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

$$SirI{coudn}^{\prime }tunderstandstep5.Howisit40?$$

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

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

Commented by ArshadS last updated on 03/Feb/25

$$\mathbb{T}\mathbf{han}\mathbb{k}\mathbf{s}\mathbb{S}\mathbf{ir}!\mathrm{I}\mathrm{understood}\mathrm{now}.$$

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