Tutor Hunt Questions
“In a triangle ABC, side AB has length 10cm , side AC has length 5cm and angle BAC = theta( measured in degrees). The area of triangle is 15cm^2
A. Find the two possible values of cos theta?
Given that BC is the longest side of the triangle,
B.find the exact length of BC.
A. Find the two possible values of cos theta?
Given that BC is the longest side of the triangle,
B.find the exact length of BC.
4 years ago
Maths Question asked by Kiran
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4 Answers
To find cosθ;;, use the formula for the area of a triangle i.e. AREA=1/2 x a x b x sinC.=> For this case: 15= 1/2 x 10 x 5 x sinC to find sinC.=> SinC = 3/5 thus, Arcsin(3/5)=+- 4/5 or +-0.8
To find the exact length of BC, use the cosine rule.=> c(sq)=a(sq)+b(sq)-2abCosC=> c(sq)=10(sq)+5(sq)-2(10)(5)(+-4/5)=> c(sq)= Square root of 205
To find the exact length of BC, use the cosine rule.=> c(sq)=a(sq)+b(sq)-2abCosC=> c(sq)=10(sq)+5(sq)-2(10)(5)(+-4/5)=> c(sq)= Square root of 205
A
Using the formula for area of the triangle we get 0.5*10*5*sin(theta)=15 so sin(theta)=3/5. Then using the identity (sinx)^2+(cos^x)^2=1, we get (3/5)^2+(cos(theta))^2=1 so (cos(theta))^2=16/25 so cos(theta)=+- 4/5
B
Using the cosine rule we get BC^2 =5^2+10^2-2*5*10*cos(theta). If we draw the two possible triangles that could be formed, we see that one gives theta being obtuse and one gives theta as being acute. We are told BC is the longest side so theta must be obtuse and the value of cos(theta) that gives this is when cos(theta)=-4/5. So BC^2=125-100(-4/5)=125+80=205. So BC is of length sqrt(205) because it is a length we took the positive root.
Using the formula for area of the triangle we get 0.5*10*5*sin(theta)=15 so sin(theta)=3/5. Then using the identity (sinx)^2+(cos^x)^2=1, we get (3/5)^2+(cos(theta))^2=1 so (cos(theta))^2=16/25 so cos(theta)=+- 4/5
B
Using the cosine rule we get BC^2 =5^2+10^2-2*5*10*cos(theta). If we draw the two possible triangles that could be formed, we see that one gives theta being obtuse and one gives theta as being acute. We are told BC is the longest side so theta must be obtuse and the value of cos(theta) that gives this is when cos(theta)=-4/5. So BC^2=125-100(-4/5)=125+80=205. So BC is of length sqrt(205) because it is a length we took the positive root.
Area of triangle = 1/2 x b x c x sin (BAC) [angle BAC = theta]
so 15 = 1/2 x 5 x 10 x sin (BAC) = 25 sin (BAC)
=> sin (BAC) = 15/25 = 3/5
trig identity: sin^2(x) + cos^2(x) = 1
=> cos^2(x) = 1 - sin^2(x)
Applying this:
cos^2(BAC) = 1 - (3/5)^2 = 16/25
=> cos(BAC) = +/- 4/5
so cos(theta) = +4/5 or -4/5 [+0.8 or -0.8]
so 15 = 1/2 x 5 x 10 x sin (BAC) = 25 sin (BAC)
=> sin (BAC) = 15/25 = 3/5
trig identity: sin^2(x) + cos^2(x) = 1
=> cos^2(x) = 1 - sin^2(x)
Applying this:
cos^2(BAC) = 1 - (3/5)^2 = 16/25
=> cos(BAC) = +/- 4/5
so cos(theta) = +4/5 or -4/5 [+0.8 or -0.8]
A. Using the formula Area = 1/2(abSinC), we can rearrange this to get SinC = 2Area/ab
Plugging in the values for Area, a and b, we get SinTheta = 3/5
Using either a right angled triangle or cos^2x + sin^2x = 1, we end up with CosTheta = 4/5 or -4/5 (if you use the right angled triangle, symmetry property of CosX gets you -4/5)
B. Using the Cosine Rule: BC^2 = AB^2 + AC^2 - 2(AB)(AC)CosTheta
Using both values for CosTheta:
CosTheta = 4/5 gets you BC = 3√;;5
CosTheta = -4/5 gets you BC = √;;205
As BC is the longest side of the triangle, BC is √;;205cm as 3√;;5 is smaller than 10.
Plugging in the values for Area, a and b, we get SinTheta = 3/5
Using either a right angled triangle or cos^2x + sin^2x = 1, we end up with CosTheta = 4/5 or -4/5 (if you use the right angled triangle, symmetry property of CosX gets you -4/5)
B. Using the Cosine Rule: BC^2 = AB^2 + AC^2 - 2(AB)(AC)CosTheta
Using both values for CosTheta:
CosTheta = 4/5 gets you BC = 3√;;5
CosTheta = -4/5 gets you BC = √;;205
As BC is the longest side of the triangle, BC is √;;205cm as 3√;;5 is smaller than 10.
