Light Houses Galore

east coast lighthouses
Can you please help me with this math question?

1. From the top T of a 135 metre cliff, the angles of depression of two cabins at A and B are 23 degrees and 42 degrees respectively. How far apart are A and B assuming that A,B and X, the foot of the cliff, are collinear?
(If its possible can you please show me how to do it specifically.

2a. A ship's captain measures the angle of elevation of a lighthouse is 105m above the sea, how far is he from the coast (to the nearest 100m)?
b. A plane files at a speed of 650 km/h. It starts from town A and flies on a bearing of 120 degrees for 3 hours. At that time, how far is it (i) south of A? (ii) east of A?

Can you please help and can you please show me how to do these questions specifically

thanks

Question 1:

i) What does Collinear mean?
It means A B and X are in a straight line

ii) How about a visual?

T--------------------------------------------------( line of sight = 0 degrees)

X B(42) A(23)

Imagine Diagonal lines drawn from T to to A and B, the angle, starting from line of sight DOWNWARD to A and B is angle of DEPRESSION

iii) Distance from T to A and T to B ?

If we "connect the dots": TXA and TXB, forms 2 triangles. We know the line of sight is 90 Degrees from X. We also know ANGLE XTA = 90-23 = 67, and similarly ANGLE XTB = 90 - 42 = 48.

We want the Length TA and TB, what we have is the distance TX (135m cliff) and the respective angles. Therefore we use the COSINE function (because TA and TB is the hypotenuses (diagonals) of the triangles).

Hopefully if you know about SOH CAH TOA, you get the following equations from that data:

COS(67) = 135m / HYP (Triangle XTA)
COS(48) = 135m / HYP (Triangle XTB)

Sloving for HYP(A) and HYP(B) gives

HYP(A) = 135 / cos(67) = 345.5m (approximately)
HYP(B) = 135 / cos(48) =201.8m (approximately)

iv) So how do we find the Distance from X to A and X to B?

We now know the hypotenuses of both triangles, therefore we can take advantage of the pathagorean theorm which states:

A^2 + B^2 = C^2 , where A and B are the sides and C is the hypothenuse (diagonal). Note ^2 means SQUARED. or the same number multiplied by iteself)

this gives the corresponding equations

(135) ^2 + B^2 = (345.5)^2 (TRIANGLE XTA)
(135) ^2 + B^2 = (201.8)^2 (TRIANGLE XTB)

Solving for the equation we get:

18,225 + B^2 = 119 370.25 (APPROXIMATELY)
B^2 = 119,370.25 - 18,225
B^2=101,145.25
B= Sqrt (101,145.25)
B=318.0 (APPROXIMATELY, for XTA)

B=150.0 (APPROXIMATELY, for XTB)

WHERE Sqrt = Square root

v) So what is the Distance from A to B?

IF X to A =318m
AND X to B = 150m
AND from part i we know that they are colinear
THEN A to B = 318 -150 = 168 m (APPROXIMATELY)

....See if you can explain #2 to me

I really wish we can draw pictures on these things, Email me and I will draw it for you in Paint if you want

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Whitley bay with its vibrant night life and Spanish city amusement park is great for Adults and kids alike, where as Tynemouth offers a more relaxing Seaside experience.

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Newcastle offers its visitors a weekend getaway to remember, with a great city centre and a seaside getaway all wrapped up in one. My advice is do not just stay in the centre, take a ride on the metro system!

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