一道数学题，请解答。

1. From a position 110km northwest of a coast guard station, an oil tanker makes radio contact with the coast guard. The tanker is travelling due south at 25km/h. The radar unit at the coast guard station has a range of 90km. For what length of time can the coast guard expect the tanker to be visible on the radar screen, to the nearest tenth of an hour?

【Solution】
Let the time needed by oil tanker travelling due south be t；
Let the distance travelled by oil tanker be D；
Let the time duration travelled by oil tanker visible on radar screen be △t.

By cosine rule, we have
D² = (25t)² + 110² - 2×25t×110×cos45°

Let D ≤ 90
So, (25t)² + 110² - 2×25t×110×cos45°≤ 90²
ie. 5t² - 22√2t + 32 ≤ 0
1.3 ≤ t ≤ 4.9

△t = 4.9 - 1.3 = 3.6 (hr)

2. A triangular garden is enclosed by a fence. A dog is on a 5m leash tethered to the fence at point P. One corner, B, of the fence is 6.5m from P and forms a 41degree angle, as shown. Determine the total length of fence that the dog can reach, to the nearest tenth of a metre, if the dog cannot reach side AC.

【Solution】
Let the distance from corner B not reached by dog be x.
By cosine rule, we can equate:
5² = x² + 6.5² -2×6.5×x×cos41°
x² - 9.81x + 17.25 = 0
x₁= 7.51 (m) , x₂= 2.30 (m)
x₁- x₂= 5.21 (m)

So, the total length of fence that the dog can reach
= 10 + 5.21 = 15.21 (m)

1.第一题，把图画下来，题目写错了，是due south“正南”

答案是“（55√2-5√80）÷25” √是根号的意思。

两个直角三角形算算就出来了

2.第二题这个角度算起来比较麻烦……

英语数学题难