In the previous section we completed a discussion on tangent of an acute angle. We also saw some solved examples. In this section we will see a few more solved examples based on the discussions that we have had so far in this chapter.
Solved example 30.7
What is the area of the ΔABC shown in fig.30.10(a) below:
Solution:
• In this problem, it is not known whether ΔABC is right angled or not
1. Area of ΔABC = 1⁄2 × Base × Altitude
2. But altitude is not given. So we drop a perpendicular AD from vertex A onto the side BC. This is shown in fig(b)
3. So now we have two right triangles: ⊿ABD and ⊿ACD
4. Consider ⊿ABD. For the angle 50o, opposite side is AD. Hypotenuse of the triangle is AB. So we can write:
sin 50 = opposite side⁄hypotenuse = AD⁄AB = AD⁄4.
5. But from the table, sin 50 = 0.7660
6. Equating (4) and (5) we get: AD⁄4 = 0.7660 ⟹ AD = 0.7660 × 4 = 3.064 cm
7. So the required area = 1⁄2 × Base × Altitude = 1⁄2 × 6 × 3.064 = 9.192 cm2
Solved example 30.8
If in the above problem, the angle at A is 130o, what would be the area?
Solution:
If the angle at A is 130, the ΔABC would be as shown in fig.30.11(a) below:
1. In this problem also, area of ΔABC = 1⁄2 × Base × Altitude
2. But altitude is not given. So we drop a perpendicular AD from vertex A onto the side BC. For that, we extend BC towards the left This is shown in fig(b)
3. So now we have two triangles: ⊿ABD and ΔABC
4. Consider ⊿ABD.The angles ∠ABD and ∠ABC are co-linear. That is., ∠ABD + ∠ABC = 180o
So we get: ∠ABD = 180 - ∠ABC = 180 - 130 = 50o
5. For this angle 50o, opposite side is AD. Hypotenuse of the triangle is AB. So we can write:
sin 50 = opposite side⁄hypotenuse = AD⁄AB = AD⁄4.
5. But from the table, sin 50 = 0.7660
6. Equating (4) and (5) we get: AD⁄4 = 0.7660 ⟹ AD = 0.7660 × 4 = 3.064 cm
7. So the required area = 1⁄2 × Base × Altitude = 1⁄2 × 6 × 3.064 = 9.192 cm2
Solved example 30.9
Find the length of the side QR of the ΔPQR in fig.30.12(a) below:
Solution:
• In this problem, it is not known whether ΔPQR is right angled or not
1. Drop a perpendicular RS from vertex R onto side PQ. Now we have two right triangles: ⊿PRS and ⊿QRS
2. Consider ⊿PRS. For the angle 40o, opposite side is RS. Hypotenuse of the triangle is PR. So we can write:
sin 40 = opposite side⁄hypotenuse = RS⁄PR = RS⁄6.
3. But from the table, sin 40 = 0.6428
4. Equating (2) and (3) we get: RS⁄6 = 0.6428 ⟹ RS = 0.6428 × 6 = 3.8568 cm
5. Consider ⊿PRS again. For the angle 40o, adjacent side is PS. Hypotenuse of the triangle is PR. So we can write:
cos 40 = adjacent side⁄hypotenuse = PS⁄PR = PS⁄6.
6. But from the table, cos 40 = 0.7660
7. Equating (5) and (6) we get: PS⁄6 = 0.7660 ⟹ PS = 0.7660 × 6 = 4.596 cm
8. Now we can find SQ:
SQ = PQ - PS = 7 - 4.596 = 2.404 cm
9. Consider ⊿QRS. Applying Pythagoras theorem, we get:
QR2 = SQ2 + SR2 ⟹ QR2 = 2.4042 + 3.85682 ⟹ QR2 = 20.654 ⟹ QR = √20.654 = 4.544 cm
Solved example 30.10
This is given in the form of a video presentation. It can be seen here
Solved example 30.11
This is given in the form of a video presentation. It can be seen here
Solved example 30.7
What is the area of the ΔABC shown in fig.30.10(a) below:
 |
| Fig.30.10 |
• In this problem, it is not known whether ΔABC is right angled or not
1. Area of ΔABC = 1⁄2 × Base × Altitude
2. But altitude is not given. So we drop a perpendicular AD from vertex A onto the side BC. This is shown in fig(b)
3. So now we have two right triangles: ⊿ABD and ⊿ACD
4. Consider ⊿ABD. For the angle 50o, opposite side is AD. Hypotenuse of the triangle is AB. So we can write:
sin 50 = opposite side⁄hypotenuse = AD⁄AB = AD⁄4.
5. But from the table, sin 50 = 0.7660
6. Equating (4) and (5) we get: AD⁄4 = 0.7660 ⟹ AD = 0.7660 × 4 = 3.064 cm
7. So the required area = 1⁄2 × Base × Altitude = 1⁄2 × 6 × 3.064 = 9.192 cm2
Solved example 30.8
If in the above problem, the angle at A is 130o, what would be the area?
Solution:
If the angle at A is 130, the ΔABC would be as shown in fig.30.11(a) below:
 |
| Fig.30.11 |
2. But altitude is not given. So we drop a perpendicular AD from vertex A onto the side BC. For that, we extend BC towards the left This is shown in fig(b)
3. So now we have two triangles: ⊿ABD and ΔABC
4. Consider ⊿ABD.The angles ∠ABD and ∠ABC are co-linear. That is., ∠ABD + ∠ABC = 180o
So we get: ∠ABD = 180 - ∠ABC = 180 - 130 = 50o
5. For this angle 50o, opposite side is AD. Hypotenuse of the triangle is AB. So we can write:
sin 50 = opposite side⁄hypotenuse = AD⁄AB = AD⁄4.
5. But from the table, sin 50 = 0.7660
6. Equating (4) and (5) we get: AD⁄4 = 0.7660 ⟹ AD = 0.7660 × 4 = 3.064 cm
7. So the required area = 1⁄2 × Base × Altitude = 1⁄2 × 6 × 3.064 = 9.192 cm2
Solved example 30.9
Find the length of the side QR of the ΔPQR in fig.30.12(a) below:
 |
| Fig.30.12 |
• In this problem, it is not known whether ΔPQR is right angled or not
1. Drop a perpendicular RS from vertex R onto side PQ. Now we have two right triangles: ⊿PRS and ⊿QRS
2. Consider ⊿PRS. For the angle 40o, opposite side is RS. Hypotenuse of the triangle is PR. So we can write:
sin 40 = opposite side⁄hypotenuse = RS⁄PR = RS⁄6.
3. But from the table, sin 40 = 0.6428
4. Equating (2) and (3) we get: RS⁄6 = 0.6428 ⟹ RS = 0.6428 × 6 = 3.8568 cm
5. Consider ⊿PRS again. For the angle 40o, adjacent side is PS. Hypotenuse of the triangle is PR. So we can write:
cos 40 = adjacent side⁄hypotenuse = PS⁄PR = PS⁄6.
6. But from the table, cos 40 = 0.7660
7. Equating (5) and (6) we get: PS⁄6 = 0.7660 ⟹ PS = 0.7660 × 6 = 4.596 cm
8. Now we can find SQ:
SQ = PQ - PS = 7 - 4.596 = 2.404 cm
9. Consider ⊿QRS. Applying Pythagoras theorem, we get:
QR2 = SQ2 + SR2 ⟹ QR2 = 2.4042 + 3.85682 ⟹ QR2 = 20.654 ⟹ QR = √20.654 = 4.544 cm
Solved example 30.10
This is given in the form of a video presentation. It can be seen here
Solved example 30.11
This is given in the form of a video presentation. It can be seen here
Solved example 30.12
This is given in the form of a video presentation. It can be seen here
This is given in the form of a video presentation. It can be seen here
In the next section we will see details of triangles with 30o, 60o and 45o.
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