Problem 1 :
A farmer wants to purchase a triangular shaped land with sides 120 feet and 60 feet and the angle included between these two sides is 60◦. If the land costs Rs. 500 per sq.ft, find the amount he needed to purchase the land. Also find the perimeter of the land.
Solution :
To find the missing side, let us use cosine formula.
a = 120 feet, b = 60 feet
The angle included by the sides a and b will be C.
<C = 60◦
cos C = (a2 + b2 - c2) / 2ab
cos 60◦ = (1202 + 602 - c2) / 2(120)(60)
1/2 = (14400 + 3600 - c2) / 14400
7200 = 18000 - c2
c2 = 18000-7200
c2 = 10800
c = 60√3
Perimeter of the triangle = 120+60+60√3
= 180 + 60√3
Area of triangle = (1/2)ab sin C
= (1/2) (120)(60) sin 60
= 60 (60)( 3/2)
= 3117.6
Cost of land = Rs 500 per square feet
Required cost = 3117.6(500)
= Rs. 1558800
Problem 2 :
A fighter jet has to hit a small target by flying a horizontal distance. When the target is sighted, the pilot measures the angle of depression to be 30◦. If after 100 km, the target has an angle of depression of 45◦, how far is the target from the fighter jet at that instant?
Solution :
From the diagram, we come to know that we have to find the value of a.
Sine formula :
a/sin A = b/sin B = c/sin C
45 + <B = 180
<B = 180 - 45 = 135
<BCA = 45 - 30 = 15
a/sin 30 = b/sin 135 = 100/sin 15
a/sin 30 = 100/sin 15
sin (45 - 30) = sin 45 cos 30 - cos 45 sin 30
= (1/√2) (√3/2) - (1/√2) (1/2)
= (√3/2√2) - (1/2√2)
= (√3 - 1)/2√2
a/(1/2) = 100/((√3 - 1)/2√2)
2a = 200√2/(√3 - 1)
a = [200√2/(√3 - 1)] [(√3 + 1)/(√3 + 1)]
= 200√2(√3 + 1)/2
a = 50√2(√3 + 1)
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