David wants to measure the width of a road
he chooses two points, A and B, 40m apart.
The angles made with a lamppost, P, on the opposite side are measured.
Angle BAP=45 degrees
Angle ABP=60 degrees
Work out the width, in metres, of the road. give your answer correct to 3 significant figures

The width of the road =  25.359 m

Step-by-step explanation:

See the attached figure which represents the problem.

The length of AB = 40 m

let the width of the road w.

Construct PC perpendicular to AB, So, the measure of angle C = 90°

∠B = 60° and ∠A = 45°

Let the length of AC = x, so, the length of BC = 40-x

At ΔBCP which is a right triangle at C

tan B = opposite/adjacent = w/(40-x)

w = (40-x) * tan B  ⇒(1)

At ΔACP which is a right triangle at C

tan A = opposite/adjacent = w/x

w = x * tan A  ⇒(2)

from (1) and (2)

(40-x) * tan B = x * tan A

40 tan B - x *tan B = x tan A

40 tan B  = x tan A + x *tan B

40 tan B  = x (tan A + tan B)

x = (40 tan B)/(tan A + tan B) = (40 tan 60)/(tan 60 + tan 45) = 25.359 m

substitute at (2) with x

w = x tan A = 25.359 tan 45 = 25.359 m

So, The width of the road =  25.359 m

2 over 5

step-by-step explanation:

the 2: 3 ratio is the length of the ac segment to the cb segment total length. this implies that the ratio between segment ac length and segment ab length is 2: 5. point c is 2/5 from a to b, so the required fraction is 2/5.

the answer is:

the answer is:

to solve the problem, we need to remember the quotient of power property, it's defined by the following relation:

if we have a quotienf of powers that have the same base, we need to keep the same base and subtract the exponent of the denominator power to the exponent of the numerator power.

so, we are given the expression:

then, calculating we have:

hence, the answer is:

have a nice day!