Option 1 is the correct answer
Step-by-step explanation:
The given figure has two right angled triangles.
On the right triangle,
Base = d
Perpendicular = b
And
Hypotenuse = H =10
So,
[tex]sin\ 60=\frac{b}{10}\\\frac{\sqrt{3}}{2}=\frac{b}{10}\\10*\frac{\sqrt{3}}{2}=b\\b=\frac{10\sqrt{3}}{2}\\b=5\sqrt{3}\\Now,\\cos\ 60=\frac{d}{10}\\\frac{1}{2}=\frac{d}{10}\\d=\frac{1}{2}*10\\d=5\\And\ the\ left\ triangle\\Perpendicular=b=5\sqrt{3}\\Base =c\\Hypotenuse=a\\So,\\sin\ 30=\frac{5\sqrt{3}}{a}\\a=\frac{5\sqrt{3}}{sin\ 30}\\a=\frac{5\sqrt{3}}{\frac{1}{2}}}\\a=2*5\sqrt{3}\\a=10\sqrt{3}\\Now, \\cos\ 30=\frac{c}{a}\\cos\ 30 = \frac{c}{10\sqrt{3}}\\[/tex]
[tex]\frac{\sqrt{3}}{2} = \frac{c}{10\sqrt{3}}\\c= \frac{\sqrt{3}}{2}*(10\sqrt{3})\\c=5 (\sqrt{3})^2\\c=5*3\\c=15[/tex]
Hence,
Option 1 is the correct answer
Keywords: Right-angled triangles, Trigonometric ratios
Learn more about geometry at:
- brainly.com/question/4639731
- brainly.com/question/4655616
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