Geometry and Trigonometry?

Question

Question: The length of two sides of a triangle are 9.3cm and 10.5cm and the included angle is 71°. Calculate:
a) the area of the triangle
b) the perimeter of the triangle

My answer:
a) Area=46.2cm^2
b) Perimeter=33.5cm (correct to 1.decimal.place)

2. A surveying team marks off two positions P and Q. They are on level ground, on the same side of a hill and in line with the hill. The horizontal distance from P to Q is 1200m and the team measure the angle of elevation of the summit of a hill from P and Q as 21° and 71° respectively. Find the height of the hill.

My answer: Height of hill=822.1m (correct to 1. decimal place).

Are the answers for question 1. a and b and question 2 correct? I have spent many hours on all of these questions. Could you please tell me if they are wrong?

Also I live in Australia...if you are not familiar with centrimetres(cm) and metres (m) here are a few conversions:
1cm=0.01metre
10cm=0.1metre (m)
100cm=1metre

1000metres=1kilometre (km

Answer 1

Well, it's nice to see you doing your work and checking. This should be in the math section instead of the physics section, but I'll let this one slide ;)

The first question involves the law of cosines.

a² = b² + c² - 2bc cos A
a² = (9.3)² + (10.5)² - 2(9.3)(10.5) cos 71º
a² = 133.157 cm²
a = 11.54 cm

Now that we know the length of all three sides, the area and perimeter is easy. To find the area, we need to use Heron's formula...

a = 11.54 cm
b = 9.3 cm
c = 10.5 cm

s = (a + b + c) / 2 = 15.67 cm

A = sqrt(s (s - a) (s - b) (s - c) )
A = 46.17 cm²

P = 31.34 cm
Your answer for area agrees with mine! Double check your addition for perimeter. It could be a rounding thing.



Part 2.
I forgot the formula that we can use, but we can deduce it locigally.

We can think of this as two triangles.

If we look at the "outside" triangle, we see it has a 21º angle. Since the other base angle is supplementary with 71º, it must be 109º. The third angle is then 50º. We want the hypotenuse of the "inner" triangle, which can be found using the Law of Sines.

sin (50º) / 1200 = sin 21º / r
r = 561.379 m

Now we can use simple trigonometry.

h = r * sin(71º)
h = 530.795m

My answer differs from yours in this case. My method is correct, and I am fairly confident in my calculations. I at least hope I gave you a starting point.

Answer 2

1) The length of two sides of a triangle are 9.3cm and 10.5cm and the included angle is 71°. Calculate:

a) the area of the triangle

The area of a triangle given two sides and the included angle is given by:

Area = (1/2)a*b*sinC = (1/2)9.3*10.5*sin(71°) ≈ 46.2 cm

b) the perimeter of the triangle

Use the Law of Cosines to find c.

c² = a² + b² - 2ab cosC = 9.3² + 10.5² - 2*9.3*10.5*cos(71°)
c² ≈ 133.15654
c ≈ 11.5 cm

Perimeter = a + b + c = 9.3 + 10.5 + 11.5 = 31.3 cm

Answer 3

some large books for those matters that I enjoyed: bigger Algebra : hall and Knight Geometry: For worry-unfastened Euclidean geometry, finding on how deep you desire to bypass, Euclid's components is the authoritative source. Trigonometry and Coordinate Geometry - no longer something comes close to to S L Loney's classics.

Answer 4

I'm not sure about first problem, it is correct when you assume that the given angle is the one which separates both given sides, but if it is not?,