In the previous section we saw some solved examples on 30o, 60o and 45o right triangles. In this section, we will see a few more solved examples.
9. Find the area of the triangle ABC shown in fig.(a) below:
Solution:
1. Draw a perpendicular CD from vertex C onto the side AB. This is shown in fig.(b)
2. ADC is a 45, 45, 90 triangle.
• AD and CD are the shortest sides. BC = CD = 10 cm
• AC is the hypotenuse
3. In a 45, 45, 90 triangle, hypotenuse is √2 times the shortest side. So we can write:
AC = √2 × CD ⟹ 8 = √2 × CD ⟹ 4 × √2 × √2 = √2 × CD ⟹ CD = 4√2
4. Area of ABC = 1⁄2 × base × altitude = 1⁄2 × AB × CD = 1⁄2 × 15 × 4√2 = 30√2 cm2.
10. Find the area of the triangle ABC shown in fig.(a) below:
Solution:
• This problem is similar to the problem 9 above. Two sides and an included angle are given.
1. Extend one of the given sides. Let us extend AC. This is shown in fig.(b)
• A perpendicular BD is dropped onto the extension
• So we get a right triangle: ⊿ABD
2. ∠BCD and ∠BCA form a linear pair. So we can write:
∠BCD + ∠BCA = 180 ⟹ ∠BCD + 120 = 180 ⟹ ∠BCD = 60o
3. Thus ⊿BCD is a 30, 60, 90 triangle. CD is the shortest side, BD is the medium side and BC is the hypotenuse
4. In a 30, 60, 90 triangle, the hypotenuse is twice the shortest side. So we can write:
BC = 2 × CD ⟹ 12 = 2 × CD ⟹ CD = 6 cm
5. In a 30, 60, 90 triangle, the medium side is √3 times the shortest side. So we can write:
BD = √3 × CD ⟹ BD = 6√3 cm
6. Area of ΔABC = 1⁄2 × base × altitude = 1⁄2 × AC × BD = 1⁄2 × 6 × 6√3 = 18√3 cm2.
11. Calculate the area of the quadrilateral ABCD shown in fig.(a) below:
Solution:
1. Three interior angles of the quadrilateral ABCD is given as 90o. So the fourth angle will also be 90o. It is a rectangle.
2. ∠DEA and ∠DEB form a linear pair. So we can write (see fig.b):
∠DEA + ∠DEB = 180 ⟹ ∠DEA + 120 = 180 ⟹ ∠DEA = 60o
3. Consider the ⊿ADE in fig.(b). It's angle at E is 60o. So ⊿ADE is a 30, 60, 90 triangle.
• AE is it's smallest side, AD is the medium side and DE is the largest side.
4. In a 30, 60, 90 triangle, the largest side will be 2 times the smallest side. So we can write:
DE = 2 × AE ⟹ 4 = 2 × AE ⟹ AE = 2 cm
5. Also in a 30, 60, 90 triangle, medium side is √3 times the smallest side. So we can write:
AD = √3 × AE ⟹ AD = √3 × 2 = 2√3
• So we have calculated the width of the rectangle. Now we have to calculate the length
6. Consider ∠ADC. We can write:
∠ADC = ∠ADE + ∠EDB + ∠BDC ⟹ 90 = 30 + ∠EDB + 30 ⟹ 90 = 60 +∠ EDB ⟹ ∠EDB = 30o
7. Now consider ΔEDB. We can write:
∠DEB + ∠EBD + ∠BDE = 180o (∵ sum of interior angles in any triangle is 180)
⟹ 120 + ∠EBD + 30 = 180 ⟹ 150 + ∠EBD = 180 ⟹ ∠EBD = 30o
8. So the base angles in ΔEDB are both 30o. It is an isosceles triangle. The sides opposite the equal angles will be of the same length
• Thus we get: EB = ED = 4 cm
9. So length of the rectangle = AB = AE + EB = 2 + 4 = 6 cm
10. Thus Area = AB × AD = 6 × 2√3 = 12√3 cm2.
12. In the fig.(a) below, AB = AC = 4 cm.
• How much is the height from A to BC?
• Find the length of BC
• What is the ratio of the sides of a triangle with angle measures 30o, 30o, 120o ?
Solution:
1. AB = AC. So ΔABC is an isosceles triangle.
• The base angles will be equal. We can write:
120 + 2 × ∠ABC = 180 ⟹ 2 ∠ABC = 60 ⟹ ∠ABC = 30o
• So we get ∠ABC = ∠ACB = 30o
2. Drop a perpendicular AD from the vertex A to the base BC. This is shown in fig.(b)
• since it is an isosceles triangle, AD will bisect BC. So we get BD = CD
3. AD is perpendicular to BC. So ∠BDA = 90o
• So BDA is a 30 60 90 triangle
♦ AD is the shortest side (opposite to the smallest angle 30)
♦ BD is the medium side (opposite to the medium angle 60)
♦ AB is the largest side (opposite to the largest angle 90)
4. In a 30, 60, 90 triangle, largest side is twice the shortest side. So we get:
• AB = 2AD ⟹ 4 = 2AD ⟹ AD = 2 cm
• So height of ΔABC = 2 cm
5. In a 30, 60, 90 triangle, medium side is √3 times the shortest side. So we get:
• BD = √3 × AD ⟹ BD = √3 × 2 = 2√3 cm
6. So length of BC = 2BD = 2 × 2√3 = 4√3 cm
7. Let AD = x. Then AB will be equal to 2x. AC will also be equal to 2x
• BD will be equal to x×√3. CD will also be equal to x×√3.
• So BC = 2 × x×√3 =2x√3
8. Now we can take ratio of sides:
• [AB : BC : AC] = [2x : 2x√3 : 2x]
Dividing by 2x we get:
• [AB : BC : AC] = [1 : 3 : 1]
13. In the fig.(a) below, DC = 1 cm.
• How much is BD?
• What is the ratio of the sides of a triangle with angle measures 45o, 60o, 75o ?
Solution:
1. Consider ⊿ADC. We have:
∠D + ∠C + ∠A = 180 ⟹ 90 + 60 + ∠A = 180 ⟹ 150 + ∠A = 180 ⟹ ∠A = 30o
2. So ⊿ADC is a 30, 60, 90 triangle. This is shown in fig.(b)
• CD is the shortest side (opposite to the smallest angle 30)
• AD is the medium side (opposite to the medium angle 60)
• AC is the largest side (opposite to the largest angle 90)
3. In a 30, 60, 90 triangle, medium side is √3 times the shortest side. So we get:
• AD = √3 × CD ⟹ BD = √3 × 1 = √3 cm
4. Consider ⊿BDA in fig.(a). We have:
∠B + ∠D + ∠A = 180 ⟹ 45 + 90 + ∠A = 180 ⟹ 135 + ∠A = 180 ⟹ ∠A = 45o
5. So BDA is a 45, 45, 90 triangle
• BD and AD are the smallest sides (opposite to the smallest angles 45)
• AB is the largest side (opposite to the largest angle 90)
6. In a 45, 45, 90 triangle, the smallest sides are equal. So we get:
BD = AD = √3 cm
7. Let CD = x cm. Then AC = 2x, AD = x × √3 cm, BD = x × √3 cm and AB = x × √2 cm
8. BC = BD + CD = x × √3 + x = x(1+√3)
9. Now we can take ratio of sides:
• [AB : BC : AC] = [x×√2 : x(1+√3) : 2x]
Dividing by x we get:
• [AB : BC : AC] = [√2 : (1+√3) : 2]
10. Note that, the angle at A = 45 + 30 = 75o
• So this is the ratio of the sides of a triangle with angle measures 45o, 60o, 75o.
14. In the fig.(a) below, how much is ∠BAD?
• Calculate the lengths AD, DC and AC.
• What is the ratio of the sides of a triangle with angle measures 15o, 45o, 120o ?
Solution:
1. ∠ADB and ∠ADC form a linear pair. So we can write:
∠ADB + ∠ADC = 180 ⟹ ∠ADB + 120 = 180 ⟹ ∠ADB = 60o.
2. Consider ⊿ADB. We have:
∠B + ∠D + ∠A = 180 ⟹ 90 + 60 + ∠A = 180 ⟹ 150 + ∠A = 180 ⟹ ∠BAD = 30o
3. So ⊿ADB is a 30 60 90 triangle. This is shown in fig.(b)
• BD is the shortest side (opposite to the smallest angle 30)
• AB is the medium side (opposite to the medium angle 60)
• AD is the largest side (opposite to the largest angle 90)
4. In a 30, 60, 90 triangle, medium side is √3 times the shortest side. So we get:
• AB = √3 × BD ⟹ AB = √3 × 1 = √3 cm
5. In a 30, 60, 90 triangle, largest side is 2 times the shortest side. So we get:
• AD = 2 × BD ⟹ AD = 2 × 1 = 2 cm
6. Consider ⊿ADC in fig.(a). We have:
∠D + ∠C + ∠A = 180 ⟹ 120 + 45 + ∠A = 180 ⟹ 165 + ∠A = 180 ⟹ ∠A = 15o
7. So the total angle at vertex A = 30 + 15 = 45
8. Now consider ΔABC. It is a 45, 45, 90 triangle
• AB and BC are the smallest sides (opposite to the smallest angles 45)
• AC is the largest side (opposite to the largest angle 90)
6. In a 45, 45, 90 triangle, the smallest sides are equal. So we get:
AB = BC = √3 cm
• So DC = BC - BD = (√3-1) cm
7. In a 45, 45, 90 triangle, the largest side is √2 times the smallest side. So we get:
AC = √2 × BC = √2 × √3 = √6
8. Now we can find the ratio
• When we are given a 15, 45, 120 triangle like ΔADC in fig(b), the perpendicular from the top vertex will fall outside the base.
• So we must extend the base towards the left. Then we can drop the perpendicular AB
9. The extended portion is BD. Let it's length be x cm
10. Then from the calculations above, we will get:
• AB = x×√3
• AD = 2x
• DC = (x×√3 - x) = x(√3-1)
• AC = x×√6
11. [AD : DC : AC] = [2x : x(√3 -1) : x×√6]
Dividing by x we get:
• [AD : DC : AC] = [2 : (√3 -1) : √6]
15. In the fig.(a) below, BD = 10 cm.
• Calculate ∠BAD and ∠BAC.
• Calculate the sides of ΔADC. Find the area of ΔADC.
• What is the ratio of the sides of a triangle with angle measures 15o, 30o, 135o ?
Solution:
1. ∠ADB and ∠ADC form a linear pair. So we can write:
∠ADB + ∠ADC = 180 ⟹ ∠ADB + 135 = 180 ⟹ ∠ADB = 45o.
2. Consider ⊿ADB. We have:
∠B + ∠D + ∠A = 180 ⟹ 90 + 45 + ∠A = 180 ⟹ 135 + ∠A = 180 ⟹ ∠BAD = 45o
3. So ⊿ADB is a 45, 45, 90 triangle. This is shown in fig.(b)
• BD and AB are the shortest sides (opposite to the smallest angle 45)
From this we get: AB = BD = 10 cm
• AD is the largest side (opposite to the largest angle 90)
4. In a 45, 45, 90 triangle, largest side is √2 times the shortest side. So we get:
• AD = √2 × BD ⟹ AD = √2 × 10 = 10√2 cm
5. Consider ⊿ADC in fig.(a). We have:
∠D + ∠C + ∠A = 180 ⟹ 135 + 30 + ∠A = 180 ⟹ 165 + ∠A = 180 ⟹ ∠A = 15o
6. So the total angle at vertex A = ∠BAC = 45 + 15 = 60o
7. Now consider ΔABC. It is a 30, 60, 90 triangle
• AB is the shortest side (opposite to the smallest angle 30)
• BC is the medium side (opposite to the medium angle 60)
• AC is the largest side (opposite to the largest angle 90)
8. In a 30, 60, 90 triangle, medium side is √3 times the shortest side. So we get:
• BC = √3 × AB ⟹ BC = √3 × 10 = 10√3 cm
♦ CD = (BC - BD) = (10√3 -10) = 10(√3-1) cm
♦ Area of ADC = 1⁄2 × base × altitude = 1⁄2 × CD × AB = 1⁄2 × 10(√3-1) × 10 = 50√2 cm2.
9. In a 30, 60, 90 triangle, largest side is 2 times the shortest side. So we get:
• AC = 2 × AB ⟹ AC = 2 × 10 = 20 cm
10. Now we can find the ratio
• When we are given a 15, 30, 135 triangle like ΔADC in fig(b), the perpendicular from the top vertex will fall outside the base.
• So we must extend the base towards the left. Then we can drop the perpendicular AB
11. The extended portion is BD. Let it's length be x cm
12. Then from the calculations above, we will get:
• AB = x
• AD = x×√2
• DC = (x×√3 - x) = x(√3-1)
• AC = 2x
13. [AD : DC : AC] = [x×√2 : x(√3 -1) : 2x]
Dividing by x we get:
• [AD : DC : AC] = [√2 : (√3 -1) : 2]
So we have completed the discussion on the special angles 30, 45, and 60. In the next section we will see two more special angles: 0o and 90o.
9. Find the area of the triangle ABC shown in fig.(a) below:
1. Draw a perpendicular CD from vertex C onto the side AB. This is shown in fig.(b)
2. ADC is a 45, 45, 90 triangle.
• AD and CD are the shortest sides. BC = CD = 10 cm
• AC is the hypotenuse
3. In a 45, 45, 90 triangle, hypotenuse is √2 times the shortest side. So we can write:
AC = √2 × CD ⟹ 8 = √2 × CD ⟹ 4 × √2 × √2 = √2 × CD ⟹ CD = 4√2
4. Area of ABC = 1⁄2 × base × altitude = 1⁄2 × AB × CD = 1⁄2 × 15 × 4√2 = 30√2 cm2.
10. Find the area of the triangle ABC shown in fig.(a) below:
Solution:
• This problem is similar to the problem 9 above. Two sides and an included angle are given.
1. Extend one of the given sides. Let us extend AC. This is shown in fig.(b)
• A perpendicular BD is dropped onto the extension
• So we get a right triangle: ⊿ABD
2. ∠BCD and ∠BCA form a linear pair. So we can write:
∠BCD + ∠BCA = 180 ⟹ ∠BCD + 120 = 180 ⟹ ∠BCD = 60o
3. Thus ⊿BCD is a 30, 60, 90 triangle. CD is the shortest side, BD is the medium side and BC is the hypotenuse
4. In a 30, 60, 90 triangle, the hypotenuse is twice the shortest side. So we can write:
BC = 2 × CD ⟹ 12 = 2 × CD ⟹ CD = 6 cm
5. In a 30, 60, 90 triangle, the medium side is √3 times the shortest side. So we can write:
BD = √3 × CD ⟹ BD = 6√3 cm
6. Area of ΔABC = 1⁄2 × base × altitude = 1⁄2 × AC × BD = 1⁄2 × 6 × 6√3 = 18√3 cm2.
11. Calculate the area of the quadrilateral ABCD shown in fig.(a) below:
1. Three interior angles of the quadrilateral ABCD is given as 90o. So the fourth angle will also be 90o. It is a rectangle.
2. ∠DEA and ∠DEB form a linear pair. So we can write (see fig.b):
∠DEA + ∠DEB = 180 ⟹ ∠DEA + 120 = 180 ⟹ ∠DEA = 60o
3. Consider the ⊿ADE in fig.(b). It's angle at E is 60o. So ⊿ADE is a 30, 60, 90 triangle.
• AE is it's smallest side, AD is the medium side and DE is the largest side.
4. In a 30, 60, 90 triangle, the largest side will be 2 times the smallest side. So we can write:
DE = 2 × AE ⟹ 4 = 2 × AE ⟹ AE = 2 cm
5. Also in a 30, 60, 90 triangle, medium side is √3 times the smallest side. So we can write:
AD = √3 × AE ⟹ AD = √3 × 2 = 2√3
• So we have calculated the width of the rectangle. Now we have to calculate the length
6. Consider ∠ADC. We can write:
∠ADC = ∠ADE + ∠EDB + ∠BDC ⟹ 90 = 30 + ∠EDB + 30 ⟹ 90 = 60 +∠ EDB ⟹ ∠EDB = 30o
7. Now consider ΔEDB. We can write:
∠DEB + ∠EBD + ∠BDE = 180o (∵ sum of interior angles in any triangle is 180)
⟹ 120 + ∠EBD + 30 = 180 ⟹ 150 + ∠EBD = 180 ⟹ ∠EBD = 30o
8. So the base angles in ΔEDB are both 30o. It is an isosceles triangle. The sides opposite the equal angles will be of the same length
• Thus we get: EB = ED = 4 cm
9. So length of the rectangle = AB = AE + EB = 2 + 4 = 6 cm
10. Thus Area = AB × AD = 6 × 2√3 = 12√3 cm2.
12. In the fig.(a) below, AB = AC = 4 cm.
• How much is the height from A to BC?
• Find the length of BC
• What is the ratio of the sides of a triangle with angle measures 30o, 30o, 120o ?
Solution:
1. AB = AC. So ΔABC is an isosceles triangle.
• The base angles will be equal. We can write:
120 + 2 × ∠ABC = 180 ⟹ 2 ∠ABC = 60 ⟹ ∠ABC = 30o
• So we get ∠ABC = ∠ACB = 30o
2. Drop a perpendicular AD from the vertex A to the base BC. This is shown in fig.(b)
• since it is an isosceles triangle, AD will bisect BC. So we get BD = CD
3. AD is perpendicular to BC. So ∠BDA = 90o
• So BDA is a 30 60 90 triangle
♦ AD is the shortest side (opposite to the smallest angle 30)
♦ BD is the medium side (opposite to the medium angle 60)
♦ AB is the largest side (opposite to the largest angle 90)
4. In a 30, 60, 90 triangle, largest side is twice the shortest side. So we get:
• AB = 2AD ⟹ 4 = 2AD ⟹ AD = 2 cm
• So height of ΔABC = 2 cm
5. In a 30, 60, 90 triangle, medium side is √3 times the shortest side. So we get:
• BD = √3 × AD ⟹ BD = √3 × 2 = 2√3 cm
6. So length of BC = 2BD = 2 × 2√3 = 4√3 cm
7. Let AD = x. Then AB will be equal to 2x. AC will also be equal to 2x
• BD will be equal to x×√3. CD will also be equal to x×√3.
• So BC = 2 × x×√3 =2x√3
8. Now we can take ratio of sides:
• [AB : BC : AC] = [2x : 2x√3 : 2x]
Dividing by 2x we get:
• [AB : BC : AC] = [1 : 3 : 1]
13. In the fig.(a) below, DC = 1 cm.
• What is the ratio of the sides of a triangle with angle measures 45o, 60o, 75o ?
Solution:
1. Consider ⊿ADC. We have:
∠D + ∠C + ∠A = 180 ⟹ 90 + 60 + ∠A = 180 ⟹ 150 + ∠A = 180 ⟹ ∠A = 30o
2. So ⊿ADC is a 30, 60, 90 triangle. This is shown in fig.(b)
• CD is the shortest side (opposite to the smallest angle 30)
• AD is the medium side (opposite to the medium angle 60)
• AC is the largest side (opposite to the largest angle 90)
3. In a 30, 60, 90 triangle, medium side is √3 times the shortest side. So we get:
• AD = √3 × CD ⟹ BD = √3 × 1 = √3 cm
4. Consider ⊿BDA in fig.(a). We have:
∠B + ∠D + ∠A = 180 ⟹ 45 + 90 + ∠A = 180 ⟹ 135 + ∠A = 180 ⟹ ∠A = 45o
5. So BDA is a 45, 45, 90 triangle
• BD and AD are the smallest sides (opposite to the smallest angles 45)
• AB is the largest side (opposite to the largest angle 90)
6. In a 45, 45, 90 triangle, the smallest sides are equal. So we get:
BD = AD = √3 cm
7. Let CD = x cm. Then AC = 2x, AD = x × √3 cm, BD = x × √3 cm and AB = x × √2 cm
8. BC = BD + CD = x × √3 + x = x(1+√3)
9. Now we can take ratio of sides:
• [AB : BC : AC] = [x×√2 : x(1+√3) : 2x]
Dividing by x we get:
• [AB : BC : AC] = [√2 : (1+√3) : 2]
10. Note that, the angle at A = 45 + 30 = 75o
• So this is the ratio of the sides of a triangle with angle measures 45o, 60o, 75o.
14. In the fig.(a) below, how much is ∠BAD?
• What is the ratio of the sides of a triangle with angle measures 15o, 45o, 120o ?
Solution:
1. ∠ADB and ∠ADC form a linear pair. So we can write:
∠ADB + ∠ADC = 180 ⟹ ∠ADB + 120 = 180 ⟹ ∠ADB = 60o.
2. Consider ⊿ADB. We have:
∠B + ∠D + ∠A = 180 ⟹ 90 + 60 + ∠A = 180 ⟹ 150 + ∠A = 180 ⟹ ∠BAD = 30o
3. So ⊿ADB is a 30 60 90 triangle. This is shown in fig.(b)
• BD is the shortest side (opposite to the smallest angle 30)
• AB is the medium side (opposite to the medium angle 60)
• AD is the largest side (opposite to the largest angle 90)
4. In a 30, 60, 90 triangle, medium side is √3 times the shortest side. So we get:
• AB = √3 × BD ⟹ AB = √3 × 1 = √3 cm
5. In a 30, 60, 90 triangle, largest side is 2 times the shortest side. So we get:
• AD = 2 × BD ⟹ AD = 2 × 1 = 2 cm
6. Consider ⊿ADC in fig.(a). We have:
∠D + ∠C + ∠A = 180 ⟹ 120 + 45 + ∠A = 180 ⟹ 165 + ∠A = 180 ⟹ ∠A = 15o
7. So the total angle at vertex A = 30 + 15 = 45
8. Now consider ΔABC. It is a 45, 45, 90 triangle
• AB and BC are the smallest sides (opposite to the smallest angles 45)
• AC is the largest side (opposite to the largest angle 90)
6. In a 45, 45, 90 triangle, the smallest sides are equal. So we get:
AB = BC = √3 cm
• So DC = BC - BD = (√3-1) cm
7. In a 45, 45, 90 triangle, the largest side is √2 times the smallest side. So we get:
AC = √2 × BC = √2 × √3 = √6
8. Now we can find the ratio
• When we are given a 15, 45, 120 triangle like ΔADC in fig(b), the perpendicular from the top vertex will fall outside the base.
• So we must extend the base towards the left. Then we can drop the perpendicular AB
9. The extended portion is BD. Let it's length be x cm
10. Then from the calculations above, we will get:
• AB = x×√3
• AD = 2x
• DC = (x×√3 - x) = x(√3-1)
• AC = x×√6
11. [AD : DC : AC] = [2x : x(√3 -1) : x×√6]
Dividing by x we get:
• [AD : DC : AC] = [2 : (√3 -1) : √6]
15. In the fig.(a) below, BD = 10 cm.
• Calculate ∠BAD and ∠BAC.
• Calculate the sides of ΔADC. Find the area of ΔADC.
• What is the ratio of the sides of a triangle with angle measures 15o, 30o, 135o ?
Solution:
1. ∠ADB and ∠ADC form a linear pair. So we can write:
∠ADB + ∠ADC = 180 ⟹ ∠ADB + 135 = 180 ⟹ ∠ADB = 45o.
2. Consider ⊿ADB. We have:
∠B + ∠D + ∠A = 180 ⟹ 90 + 45 + ∠A = 180 ⟹ 135 + ∠A = 180 ⟹ ∠BAD = 45o
3. So ⊿ADB is a 45, 45, 90 triangle. This is shown in fig.(b)
• BD and AB are the shortest sides (opposite to the smallest angle 45)
From this we get: AB = BD = 10 cm
• AD is the largest side (opposite to the largest angle 90)
4. In a 45, 45, 90 triangle, largest side is √2 times the shortest side. So we get:
• AD = √2 × BD ⟹ AD = √2 × 10 = 10√2 cm
5. Consider ⊿ADC in fig.(a). We have:
∠D + ∠C + ∠A = 180 ⟹ 135 + 30 + ∠A = 180 ⟹ 165 + ∠A = 180 ⟹ ∠A = 15o
6. So the total angle at vertex A = ∠BAC = 45 + 15 = 60o
7. Now consider ΔABC. It is a 30, 60, 90 triangle
• AB is the shortest side (opposite to the smallest angle 30)
• BC is the medium side (opposite to the medium angle 60)
• AC is the largest side (opposite to the largest angle 90)
8. In a 30, 60, 90 triangle, medium side is √3 times the shortest side. So we get:
• BC = √3 × AB ⟹ BC = √3 × 10 = 10√3 cm
♦ CD = (BC - BD) = (10√3 -10) = 10(√3-1) cm
♦ Area of ADC = 1⁄2 × base × altitude = 1⁄2 × CD × AB = 1⁄2 × 10(√3-1) × 10 = 50√2 cm2.
9. In a 30, 60, 90 triangle, largest side is 2 times the shortest side. So we get:
• AC = 2 × AB ⟹ AC = 2 × 10 = 20 cm
10. Now we can find the ratio
• When we are given a 15, 30, 135 triangle like ΔADC in fig(b), the perpendicular from the top vertex will fall outside the base.
• So we must extend the base towards the left. Then we can drop the perpendicular AB
11. The extended portion is BD. Let it's length be x cm
12. Then from the calculations above, we will get:
• AB = x
• AD = x×√2
• DC = (x×√3 - x) = x(√3-1)
• AC = 2x
13. [AD : DC : AC] = [x×√2 : x(√3 -1) : 2x]
Dividing by x we get:
• [AD : DC : AC] = [√2 : (√3 -1) : 2]
So we have completed the discussion on the special angles 30, 45, and 60. In the next section we will see two more special angles: 0o and 90o.
No comments:
Post a Comment