Question

to divide a line segment PQ in the ratio 7:3 internally, first a ray PX is drawn so that angle PQR is an acute angle and then at equal distance, points are marked on the ray P such that the minimum number of these points is?

Answer 1

զนεรtıσŋ

to divide a line segment PQ in the ratio 7:3 internally, first a ray PX is drawn so that angle PQR is an acute angle and then at equal distance, points are marked on the ray P such that the minimum number of these points is?

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ลиรωэя

to divide a line segment AB in the ratio ,m : n, a ray AX making an acute ∠BAX, is drawn and then m + n points are marked at equal distances on the ray AX.

Here, m = 5, n = 7

Therefore, minimum number of points to be marked on AX = m + n = 5 + 7 = l2

❤

Answer 2

Answer:

12 simple just add it bro