the angle of elevation of top of the tower from two points at the distance of 4 m and 9 M from the base of Tower and in the same straight line with 8 are complement .Prove that the height of the tower is 6m
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HELLO DEAR,
For angle of elevation A, from one point, it is (90- A) from the other point. (Since angles are complementary.)
We use tanA = height of tower/4
And tan (90-A) = height of tower/9
When we equate, the above, we get:
tanA x tan(90-A) = (height)2(height)2 / 4 x 9
Since tan(90-A) = cotA.
And, tanA x cot A = 1
Hence, height = √4√4 x √9√9= 6 m
For angle of elevation A, from one point, it is (90- A) from the other point. (Since angles are complementary.)
We use tanA = height of tower/4
And tan (90-A) = height of tower/9
When we equate, the above, we get:
tanA x tan(90-A) = (height)2(height)2 / 4 x 9
Since tan(90-A) = cotA.
And, tanA x cot A = 1
Hence, height = √4√4 x √9√9= 6 m
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