Sin(α)=opposite leg/hypotenuse
Sin(63°)=ya/35
ya=35xSin(63°)
ya=31.2
Area(a)= (15.9x31.2)/2
Area(a)=248.0
Now, we proceed to find the area of "b". We already have the length x=15.9, so:
Tan(α)=opposite leg/adjacent leg
yb=15.90xTan(42°)
yb=14.3
Area(b)=15.9X14.3/2
Area(b)=113.7
The total Area (At) is:
At=Area(a)+Area(b)
At=248.0+113.7
At=361.7
10) Triangle "a":
-Adjacent leg of the triangle "a":
Cos(α)=adjacent leg/hypotenuse
Cos(30°)=xa/48
xa=48xCos(30°)
xa=41.6
-Opposite leg of the triangle "a":
Sin(α)=opposite leg/hypotenuse
Sin(30°)=y/48
y=48xSin(30°)
y=24.0
-Area of the triangle "a":
Area(a)=41.6x24.0/2
Area(a)=499.2
Triangle "b":
-We have the value of i(y=24).
-Adjacent leg of the triangle "b":
Tan(α)=opposite leg/adjacent leg
Tan(45°)=24/xb
xb=24/Tan(45°)
xb=24
-Area of the triangle "b":
Area(b)= 24x24/2
Area(b)=288
-Total area of the triangle (At) is:
At=Area(a)+Area(b)
At=499.2+288
At=787.2
11) To find the area of the triangle shown in this exercise, we have to apply the same procedure as in exercise 10:
-Adjacent leg of the triangle "a":
Cos(α) = adjacent leg/hypotenuse
Cos(56°)=xa/14
xa=14xCos(56°)
xa=7.8
-Opposite leg of the triangle "a":
Sin(α)=opposite leg/hypotenuse
Sin(56°)=y/14
y=14xSin(56°)
y=11.6
-Area of the triangle "a":
Area(a)=7.8x11.6/2
Area(a)=45.2
Triangle "b":
-The value of its opposite leg is y=11.6.
-Adjacent leg of the triangle "b":
Tan(α)=opposite leg/adjacent leg
Tan(46°)=11.6/xb
xb=11.6/Tan(46°)
xb=11.2
-Area of the triangle "b":
Area(b)= 11.6x11.2/2
Area(b)=65.0
-Total area of the triangle (At) is:
At=Area(a)+Area(b)
At=45.2+65.0
At=110.2
12)-Adjacent leg of the triangle "a":
Cos(α)=adjacent leg/hypotenuse
Cos(54°)=xa/13
xa=13xCos(54°)
xa=7.6
-Opposite leg of the triangle "a":
Sin(α)=opposite leg/hypotenuse
Sin(54°)=y/13
y=13xSin(54°)
y=10.5
-Area of the triangle "a":
Area(a)=7.6x10.5/2
Area(a)=39.9
Triangle "b":
-Adjacent leg of the triangle "b":
Tan(α)=opposite leg/adjacent leg
Tan(42°)=7.6/xb
xb=7.6/Tan(42°)
xb=8.4
-Area of the triangle "b":
Area(b)=7.6x8.4/2
Area(b)=31.9
-Total area of the triangle (At) is:
At=Area(a)+Area(b)
At=39.9+31.9
At=71.8First, it is important to remember that the formula to calculate the area of a triangle is: A=bxh/2
As we can see in the exercises, all the triangles are divided in two triangles. So, let's call "a" to the triangle on the left and "b" to the triangle on the right.
9)To find the area of the triangle "a", we need the lenght of the adjacent leg (x) and the opposite leg (ya):
Cos(α)=adjacent leg/hypotenuse
Cos(63°)=x/35
x=35xCos(63°)
x=15.9
Sin(α)=opposite leg/hypotenuse
Sin(63°)=ya/35
ya=35xSin(63°)
ya=31.2
Area(a)= (15.9x31.2)/2
Area(a)=248.0
Now, we proceed to find the area of "b". We already have the length x=15.9, so:
Tan(α)=opposite leg/adjacent leg
yb=15.90xTan(42°)
yb=14.3
Area(b)=15.9X14.3/2
Area(b)=113.7
The total Area (At) is:
At=Area(a)+Area(b)
At=248.0+113.7
At=361.7
10) Triangle "a":
-Adjacent leg of the triangle "a":
Cos(α)=adjacent leg/hypotenuse
Cos(30°)=xa/48
xa=48xCos(30°)
xa=41.6
-Opposite leg of the triangle "a":
Sin(α)=opposite leg/hypotenuse
Sin(30°)=y/48
y=48xSin(30°)
y=24.0
-Area of the triangle "a":
Area(a)=41.6x24.0/2
Area(a)=499.2
Triangle "b":
-We have the value of i(y=24).
-Adjacent leg of the triangle "b":
Tan(α)=opposite leg/adjacent leg
Tan(45°)=24/xb
xb=24/Tan(45°)
xb=24
-Area of the triangle "b":
Area(b)= 24x24/2
Area(b)=288
-Total area of the triangle (At) is:
At=Area(a)+Area(b)
At=499.2+288
At=787.2
11) To find the area of the triangle shown in this exercise, we have to apply the same procedure as in exercise 10:
-Adjacent leg of the triangle "a":
Cos(α) = adjacent leg/hypotenuse
Cos(56°)=xa/14
xa=14xCos(56°)
xa=7.8
-Opposite leg of the triangle "a":
Sin(α)=opposite leg/hypotenuse
Sin(56°)=y/14
y=14xSin(56°)
y=11.6
-Area of the triangle "a":
Area(a)=7.8x11.6/2
Area(a)=45.2
Triangle "b":
-The value of its opposite leg is y=11.6.
-Adjacent leg of the triangle "b":
Tan(α)=opposite leg/adjacent leg
Tan(46°)=11.6/xb
xb=11.6/Tan(46°)
xb=11.2
-Area of the triangle "b":
Area(b)= 11.6x11.2/2
Area(b)=65.0
-Total area of the triangle (At) is:
At=Area(a)+Area(b)
At=45.2+65.0
At=110.2
12)-Adjacent leg of the triangle "a":
Cos(α)=adjacent leg/hypotenuse
Cos(54°)=xa/13
xa=13xCos(54°)
xa=7.6
-Opposite leg of the triangle "a":
Sin(α)=opposite leg/hypotenuse
Sin(54°)=y/13
y=13xSin(54°)
y=10.5
-Area of the triangle "a":
Area(a)=7.6x10.5/2
Area(a)=39.9
Triangle "b":
-Adjacent leg of the triangle "b":
Tan(α)=opposite leg/adjacent leg
Tan(42°)=7.6/xb
xb=7.6/Tan(42°)
xb=8.4
-Area of the triangle "b":
Area(b)=7.6x8.4/2
Area(b)=31.9
-Total area of the triangle (At) is:
At=Area(a)+Area(b)
At=39.9+31.9
At=71.8
Step-by-step explanation:
