| |
|
|
| |
 |
| Solution of Right Angled Triangles |
|
| To solve a right angled triangle, we need to find out the unknown sides and the angles with the help of t-ratios. |
| |
| To find a side, usually we take such t-ratios that involve the unknown sides. |
| |
 |
| |
 |
| |
In the figure,
 AB = 100 cm, find (i) x and (ii) y. |
| |
 |
| |
 |
| |
 |
| |
 |
| |
 x = 0.7660 x 100 |
| |
| = 76.60 cm |
| |
 |
| |
 |
| |
 y = 0.6428 x 100 |
| |
| = 64.28 cm. |
| |
 |
| |
In the figure,
 AC = 100 m,
 find (i)
 (ii) BC. Give your answer correct to the nearest metre. |
| |
 |
| |
|
|
| |
 90o  89o 60' |
| |
| 44o 36' |
| |
| 45o 24' |
| |
| = 45o 24' |
| |
 |
| |
 |
| |
 1.0141  |
| |
 BC = 100 x 1.0141 |
| |
 BC = 101.41 |
| |
 |
| |
| We have taken ÐA because, the side opposite to A is the unknown side. By this simple observation, we have done multiplication instead of division during calculations. |
| |
| To find height of a Triangle |
| |
| (a) When the altitude falls within the base. |
| |
| In DABC, AD is its height. Let AD = h and BC = a. |
| |
 |
| |
Prove that
 |
| |
In DABC,  |
| |
  |
| |
In DACD,  |
| |
  |
| |
 BC = BD + DC  |
| |
 |
| |
  |
| |
  |
| |
| (b) When the altitude falls outside the base |
| |
| In DABC,
ÐACB is obtuse. |
| |
 |
| |
| BC = a and AD = h. |
| |
| Proceeding as in (a) above, we can prove that |
| |
  |
| |
 |
| |
| In DABC,
ÐB = 42o 36', ÐC
= 50o 12' BC = 100 m. Find AD. |
| |
| |
Using  (where a is the base) |
| |
 |
| |
 |
| |
 |
| |
 |
| |
 |
| |
 |
| |
| = 52.05 |
| |
 |
| |
| In the figure, ÐACD
= 50o, ÐB = 33o 54' BC = 100 m. Find AD. |
| |
 |
| |
|
|
| |
 |
| |
 |
| |
 |
| |
 |
| |
 |
| |
| AD = 154 m |
| |
|
|
| |
|
|
| |
|
|
|
|
|
Get FREE Live Tutoring
(No credit card required)
Customer Care
Click to get customer service, technical support and subscription help.
Refer-A-Friend
Get One Month Free!
When you refer a friend
Solution of Right Angled Triangles
|
|
|
