Answer:
The point C is 12.68 km away from the point A on a bearing of S23.23°W.
Step-by-step explanation:
Given that AB is 50 km and BC is 40 km as shown in the figure.
From the figure, the length of x-component of AC = |AB sin 80° - BC cos 20°|
=|50 sin 80° - 40 cos 20°|=11.65 km
The length of y-component of AC = |AB cos 80° - BC sin 20°|
=|50 cos 80° - 40 sin 20°|= 5 km
tan
= 5/11.65
=23.23°
AC=
km
Hence, the point C is 12.68 km away from the point A on a bearing of S23.23°W.
(3 points) The altitude of a triangle is increasing at a rate of 1 cm/min<span> while the area of the triangle is increasing at a rate of 20 cm2/min. At what rate is the base of the triangle changing when the altitude is 10 cm and the area is 100 cm2. = 2, namely, the base of the triangle is increasing at a rate of </span>2 cm/min<span>.</span>
Answer:
45 cm
Step-by-step explanation:
![l=\sqrt{r^2+h^2}=\sqrt{27^2+36^2} =45 cm](https://tex.z-dn.net/?f=l%3D%5Csqrt%7Br%5E2%2Bh%5E2%7D%3D%5Csqrt%7B27%5E2%2B36%5E2%7D%20%3D45%20cm)