Q=Two trees,A and B are on the same side of a river .From a point C in the river the distance of the trees A and B is 250 m and 300 m,respectively .If the angle C is 45o ,find the distance b/w the trees (use root 2 =1.44). PLS CALCULATE AND SHOW
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Simple application of Pythagoras theorem.
see diagram.
Draw a perpendicular AD from A on to BC. In the right angle triangle ΔACD,
∠ACD = 45° ∠D=90° So ∠CAD = 45°
Hence, AD= CD.
As AC² = AD² + CD²,
So AD = CD = AC/√2 = 250/√2 = 125√2 cm = 180 m
So BD = 300 - 180 = 120 m
In the right angle ΔADB, AB² = AD² + DB²
AB² = 180² + 120²
AB = 60 √13 m
see diagram.
Draw a perpendicular AD from A on to BC. In the right angle triangle ΔACD,
∠ACD = 45° ∠D=90° So ∠CAD = 45°
Hence, AD= CD.
As AC² = AD² + CD²,
So AD = CD = AC/√2 = 250/√2 = 125√2 cm = 180 m
So BD = 300 - 180 = 120 m
In the right angle ΔADB, AB² = AD² + DB²
AB² = 180² + 120²
AB = 60 √13 m
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kvnmurty:
click on red heart thanks above pls
wow !! great explanation.
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