Question

Please help with question below

Answer 1

Answer:

○ a) $\displaystyle 9\:m.$

Step-by-step explanation:

$\displaystyle \frac{15}{x} = sec\:53 \hookrightarrow xsec\:53 = 15 \hookrightarrow \frac{15}{sec\:53} = x \hookrightarrow 9,0272253473... = x \\ \\ 9 ≈ x$

OR

$\displaystyle \frac{x}{15} = cos\:53 \hookrightarrow 15cos\:53 = x \hookrightarrow 9,0272253473... = x \\ \\ 9 ≈ x$

Information on trigonometric ratios

$\displaystyle \frac{OPPOCITE}{HYPOTENUSE} = sin\:θ \\ \frac{ADJACENT}{HYPOTENUSE} = cos\:θ \\ \frac{OPPOCITE}{ADJACENT} = tan\:θ \\ \frac{HYPOTENUSE}{ADJACENT} = sec\:θ \\ \frac{HYPOTENUSE}{OPPOCITE} = csc\:θ \\ \frac{ADJACENT}{OPPOCITE} = cot\:θ$

I am joyous to assist you at any time.

Answer 2

Hypotenuse = 15

Adjacent =x

H + A = cah or cos

Cos(53) = x/15

x15

15 x Cos(53) = x

x = 9.0272...

Ans: A) 9.0m

Hope this helps!