Question-209639

Question Number 209639 by SonGoku last updated on 17/Jul/24

Commented by SonGoku last updated on 17/Jul/24

In the figure, DC / / FG is the width of a river .

How wideis this river ?

(How do I solve it ?)

Answered by mr W last updated on 17/Jul/24

Commented by SonGoku last updated on 17/Jul/24

You are fantastic!

I learn a lot from your elegantt soluions .

Thank you very much .

Commented by SonGoku last updated on 17/Jul/24

Did you use the law of sines ?

\frac{y}{\sin 17 ° 0^{'} 50^{″}} = \frac{50}{\sin 90 °}

Commented by mr W last updated on 17/Jul/24

y = w i d t h o f r i v e r

y = 50 \sin α

\cos α = \frac{2 . 9^{2} + 1 . 3^{2} - 1 . 7^{2}}{2 \times 2.9 \times 1.3} = 0.956233

\Rightarrow \sin α = \sqrt{1 - 0 . 956233^{2}} = 0.292605

\Rightarrow y = 50 \times 0.292605 = 14.63 m

Commented by mr W last updated on 17/Jul/24

i n t h e r i g h t t r i a n g l e :

\frac{y}{50} = \sin α

y = 50 \times \sin α

Commented by mr W last updated on 17/Jul/24

Commented by SonGoku last updated on 17/Jul/24

It is always an honor to learn from you .

My sincereh tanks .

Commented by mr W last updated on 17/Jul/24

t h a n k s t o y o u t o o!

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