Advertisements
Advertisements
प्रश्न
If each angle of a triangle is less than the sum of the other two angles of it; prove that the triangle is acute-angled.
Advertisements
उत्तर
Consider ΔABC.
Now, ∠A < ∠B + ∠C
⇒ ∠A + ∠A < ∠A + ∠B + ∠C
⇒ 2∠A < 180°
⇒ ∠A < `(180°)/(2)`
⇒ ∠A < 90°
Similarly, we have
∠B < 90° and ∠C < 90°.
Hence, the triangle is acute-angled.
APPEARS IN
संबंधित प्रश्न
In the given figure, ∠Q: ∠R = 1: 2. Find:
a. ∠Q
b. ∠R
The exterior angles, obtained on producing the side of a triangle both ways, are 100° and 120°. Find all the angles of the triangle.
Use the given figure to find the value of x in terms of y. Calculate x, if y = 15°.
In a triangle PQR, ∠P + ∠Q = 130° and ∠P + ∠R = 120°. Calculate each angle of the triangle.
The angles of a triangle are (x + 10)°, (x + 30)° and (x - 10)°. Find the value of 'x'. Also, find the measure of each angle of the triangle.
In the figure given below, if RS is parallel to PQ, then find the value of ∠y.
Use the given figure to show that: ∠p + ∠q + ∠r = 360°.
If bisectors of angles A and D of a quadrilateral ABCD meet at 0, then show that ∠B + ∠C = 2 ∠AOD
In a triangle, the sum of two angles is 139° and their difference is 5°; find each angle of the triangle.
In a right-angled triangle ABC, ∠B = 90°. If BA and BC produced to the points P and Q respectively, find the value of ∠PAC + ∠QCA.
