In Δ ABC, ∠B = 90º, AC = 29 cm and BC = 20 cm. Then \(\frac{{1 - \sin A + \cos A}}{{1 + \sin A + \cos A}}\) is equal to:
3/7
Given:
In the given triangle Δ ABC, ∠B = 90°, AC = 29 cm, BC = 20 cm.
Calculation:
As shown above, in triangle Δ ABC where AB, BC, and AC are given. Since (21, 20, 29) are Pythagorean triplets, AB is 21 cm.
Now, the values of sinA and cosA can be easily calculated from the figure above.
⇒ SinA = (Opposite)/(Hypotenuse) = BC/AC = 20/29
⇒ CosA = (Adjacent)/(Hypotenuse) = AB/AC = 21/29
Now, calculating (1 – sinA + cosA)/(1 + sinA + cosA) will yield the value
⇒ (1 – 20/29 + 21/29)/(1 + 20/29 + 21/29)
⇒ (1 + 1/29)/(1 + 41/29) = (30/29)/(70/29)
⇒ 30/70 = 3/7
∴ The value of (1 – sinA + cosA)/(1 + sinA + cosA) is 3/7.
Important Questions from Properties of Triangles
Consider the following statements:
1. If ABC is a right-angled triangle, right-angled at A, and if sin \(\rm B = \frac 1 3,\) then cosec C = 3.
2. If b cos B = c cos C and if the triangle ABC is not right-angled, then ABC must be isosceles.
Which of the above statements is/are correct?
- In a triangle \(ABC\), if \(a\), \(b\) and \(c\) are the lengths of the sides opposite to the angles \(A\), \(B\) and \(C\) respectively, then what is \(\frac{\sin(A-B)}{\sin(A+B)}\) equal to ?
In equilateral ΔABC, D and E are points on the sides AB and AC, respectively, such that AD = CE. BE and CD intersect at F. The measure (in degrees) of ∠CFB is:
The sides of a triangle are m, n and \(\rm \sqrt{m^2+n^2+mn}\) . What is the sum of the acute angles of the triangle?
Each side of a square subtends an angle of 60° at the tip of a tower of height h meters standing at the centre of the square. If l is the length of each side of the square, then what is h 2equal to?
