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| Solution of Right Angled Triangles |
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| To solve a right angled triangle, we need to find out the unknown sides and the angles with the help of t-ratios. |
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| To find a side, usually we take such t-ratios that involve the unknown sides. |
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In the figure,
AB = 100 cm, find (i) x and (ii) y. |
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x = 0.7660 x 100 |
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| = 76.60 cm |
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y = 0.6428 x 100 |
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| = 64.28 cm. |
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In the figure,
AC = 100 m,
find (i)
(ii) BC. Give your answer correct to the nearest metre. |
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90o 89o 60' |
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| 44o 36' |
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| 45o 24' |
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| = 45o 24' |
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1.0141  |
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BC = 100 x 1.0141 |
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BC = 101.41 |
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| 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. |
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| To find height of a Triangle |
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| (a) When the altitude falls within the base. |
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| In DABC, AD is its height. Let AD = h and BC = a. |
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Prove that
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In DABC, |
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In DACD,  |
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BC = BD + DC  |
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| (b) When the altitude falls outside the base |
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| In DABC,
ÐACB is obtuse. |
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| BC = a and AD = h. |
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| Proceeding as in (a) above, we can prove that |
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| In DABC,
ÐB = 42o 36', ÐC
= 50o 12' BC = 100 m. Find AD. |
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Using (where a is the base) |
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| = 52.05 |
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| In the figure, ÐACD
= 50o, ÐB = 33o 54' BC = 100 m. Find AD. |
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| AD = 154 m |
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Solution of Right Angled Triangles
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