(a) The angle of depression of a boat from the mid-point of a vertical cliff is 35°. If the boat is 120m from the foot of the cliff, calculate the height of the cliff.
(b) Towns P and Q are x km apart. Two motorists set out at the same time from P to Q at steady speeds of 60 km/h and 80 km/h. The faster motorist got to Q 30 minutes earlier than the other. Find the value of x.
(b) Towns P and Q are x km apart. Two motorists set out at the same time from P to Q at steady speeds of 60 km/h and 80 km/h. The faster motorist got to Q 30 minutes earlier than the other. Find the value of x.
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Correct Answer: Option
Explanation:
(a)
\(\frac{FM}{FB} = \tan 35\)
\(\therefore \frac{FM}{120} = \tan 35\)
\(FM = 120 \tan 35 \)
= \(120 \times 0.7002\)
= \(84.024 m\)
Therefore, the height of the cliff = \(2 \times 84.024 = 168.048m\)
(b) Distance between P and Q = x km
Speed for 1st motorist = 60 km/h
\(\therefore \text{1st motorist's time} = \frac{x}{60} hrs\)
Speed for 2nd motorist = 80 km/h
\(\therefore \text{2nd motorist's time} = \frac{x}{80} hrs\)
\(\frac{x}{60} - \frac{x}{80} = \frac{30}{60}\)
\(\frac{4x - 3x}{240} = \frac{1}{2}\)
\(\frac{x}{240} = \frac{1}{2} \implies 2x = 240\)
\(x = 120 km\).
(a)
\(\frac{FM}{FB} = \tan 35\)
\(\therefore \frac{FM}{120} = \tan 35\)
\(FM = 120 \tan 35 \)
= \(120 \times 0.7002\)
= \(84.024 m\)
Therefore, the height of the cliff = \(2 \times 84.024 = 168.048m\)
(b) Distance between P and Q = x km
Speed for 1st motorist = 60 km/h
\(\therefore \text{1st motorist's time} = \frac{x}{60} hrs\)
Speed for 2nd motorist = 80 km/h
\(\therefore \text{2nd motorist's time} = \frac{x}{80} hrs\)
\(\frac{x}{60} - \frac{x}{80} = \frac{30}{60}\)
\(\frac{4x - 3x}{240} = \frac{1}{2}\)
\(\frac{x}{240} = \frac{1}{2} \implies 2x = 240\)
\(x = 120 km\).