I have problem with solving question 19 and 24th of the link below.
http://www.lhvarsitymath.com/Trig%20Word…
The answers are already there but I need the steps I tried but could not get the same answers for both. Can you go to the link and show me the steps? I appreciate your help in any of the questions. Thanks.
http://www.lhvarsitymath.com/Trig%20Word…
The answers are already there but I need the steps I tried but could not get the same answers for both. Can you go to the link and show me the steps? I appreciate your help in any of the questions. Thanks.
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19) The angles A and B are, respectively,
(90-12)° = 78° and (90-30)° = 60°
Applying the sine rule, the distance of the ship x mile from A is given by
x/sin60° = 2/sin42°
x = 2.588509
i.e. it's about 2.6 mile.
The distance y mile from B is given by
y/sin78° = 2/sin42°
y = 2.923637
i.e. about 2.9 mile
The distance h mile from shore is
h = x cos12° = y cos 30°
.. = 2.531944
i.e. about 2.5 mile
24) The diagonal top left to bottom right divides the quadrilateral into two triangles. The area of one of them is
(1/2)*50*100*sin40° sq ft
= 1606.969 sq ft to three decimal places
The only way I can see to calculate the other is to calculate the diagonal d and then use the cosine rule in the other triangle to find the remaining side of the quadrilateral, and then we can calculate the area of that other triangle.
d² = 50² + 100² - 2*50*100*cos40°
d = 69.566914
then
d² = 20² + x² - 2*20*x*cos100°
x² - (40cos100°)x + 20² - d² = 0
x = [40*cos100° ±√((40*cos100°)² - 4*(20² - d²))]/2
One root is negative, and we want the positive root:
x = 63.247476 to six decimal places
Area of triangle = (1/2)*20*x*sin100°
....................... = 622.866 sq ft to 3 decimal places.
Total area of quadrilateral = ((1/2)*50*100*sin40° + (1/2)*20*x*sin100°) sq ft
........................................ = 2229.8 sq ft correct to 1 decimal place
(90-12)° = 78° and (90-30)° = 60°
Applying the sine rule, the distance of the ship x mile from A is given by
x/sin60° = 2/sin42°
x = 2.588509
i.e. it's about 2.6 mile.
The distance y mile from B is given by
y/sin78° = 2/sin42°
y = 2.923637
i.e. about 2.9 mile
The distance h mile from shore is
h = x cos12° = y cos 30°
.. = 2.531944
i.e. about 2.5 mile
24) The diagonal top left to bottom right divides the quadrilateral into two triangles. The area of one of them is
(1/2)*50*100*sin40° sq ft
= 1606.969 sq ft to three decimal places
The only way I can see to calculate the other is to calculate the diagonal d and then use the cosine rule in the other triangle to find the remaining side of the quadrilateral, and then we can calculate the area of that other triangle.
d² = 50² + 100² - 2*50*100*cos40°
d = 69.566914
then
d² = 20² + x² - 2*20*x*cos100°
x² - (40cos100°)x + 20² - d² = 0
x = [40*cos100° ±√((40*cos100°)² - 4*(20² - d²))]/2
One root is negative, and we want the positive root:
x = 63.247476 to six decimal places
Area of triangle = (1/2)*20*x*sin100°
....................... = 622.866 sq ft to 3 decimal places.
Total area of quadrilateral = ((1/2)*50*100*sin40° + (1/2)*20*x*sin100°) sq ft
........................................ = 2229.8 sq ft correct to 1 decimal place