SSC (English Medium) 10th Standard - Maharashtra State Board Question Bank Solutions for Geometry Mathematics 2

Construct an equilateral ∆ABC with side 5 cm. ∆ABC ~ ∆LMN, ratio the corresponding sides of triangle is 6 : 7, then construct ΔLMN and ΔABC.

ΔAMT ~ ΔAHE. In ΔAMT, AM = 6.3 cm, ∠MAT = 120°, AT = 4.9 cm, `(AM)/(HA) = 7/5`, then construct ΔAMT and ΔAHE.

ΔRHP ~ ΔNED, In ΔNED, NE = 7 cm, ∠D = 30°, ∠N = 20°, `(HP)/(ED) = 4/5`, then construct ΔRHP and ∆NED.

ΔABC ~ ΔPBR, BC = 8 cm, AC = 10 cm, ∠B = 90°, `(BC)/(BR) = 5/4` then construct ∆ABC and ΔPBR.

If the point P(6, 7) divides the segment joining A(8, 9) and B(1, 2) in some ratio, find that ratio.

Solution:

Point P divides segment AB in the ratio m : n.

A(8, 9) = (x₁, y₁), B(1, 2) = (x₂, y₂) and P(6, 7) = (x, y)

Using Section formula of internal division,

∴ `7 = (m(square) - n(9))/(m + n)`

∴ 7m + 7n = `square` + 9n

∴ 7m – `square` = 9n – `square`

∴ `square` = 2n

∴ `m/n = square`

cos θ . sec θ = ?

sec 60° = ?

1 + cot²θ = ? 

cot θ . tan θ = ?

sec²θ – tan²θ = ?

sin²θ + sin²(90 – θ) = ?

`(1 + cot^2A)/(1 + tan^2A)` = ?

`sin θ = 1/2`, then θ = ?

tan (90 – θ) = ?

cos 45° = ?

Which is not correct formula?

`(1 - tan^2 45^circ)/(1 + tan^2 45^circ)` = ?

If `tan θ = 13/12`, then cot θ = ?

Prove that `"cosec"  θ xx sqrt(1 - cos^2θ) = 1`.

If `1 - cos^2θ = 1/4`, then θ = ?