Advertisements
Advertisements
प्रश्न
Find AB.

Advertisements
उत्तर
Consider the figure

From right triangle ACF
tan 45° = `(20)/"AC"`
1 = `(20)/"AC"`
AC = 20 cm
From triangle DEB
tan 60° = `(30)/"BD"`
`sqrt(3) = (30)/"BD"`
BD = `(30)/sqrt(3)` = 17.32 cm
Given FC = 20, ED = 30, So EP = 10 cm
Therefore
tan 60° = `"FP"/"EP"`
`sqrt(3)= "FP"/(10)`
FP = `10sqrt(3)` = 17.32 cm
Thus AB = AC + CD + BD = 54.64 cm.
APPEARS IN
संबंधित प्रश्न
Find ‘x’, if:

Find 'x', if :
Find angle 'A' if :
Find angle 'A' if:

Find AD, if :
In trapezium ABCD, as shown, AB // DC, AD = DC = BC = 20 cm and A = 60°. Find: distance between AB and DC.

In right-angled triangle ABC; ∠ B = 90°. Find the magnitude of angle A, if: AB is √3 times of BC.
In right-angled triangle ABC; ∠B = 90°. Find the magnitude of angle A, if:
BC is `sqrt(3)` times of AB.
Find PQ, if AB = 150 m, ∠ P = 30° and ∠ Q = 45°.

The perimeter of a rhombus is 96 cm and obtuse angle of it is 120°. Find the lengths of its diagonals.
