Advertisements
Advertisements
प्रश्न
In the given figure, ∠Q: ∠R = 1: 2. Find:
a. ∠Q
b. ∠R
Advertisements
उत्तर
∠Q : ∠R = 1 : 2
Let ∠Q = x°
⇒ ∠R = 2x°
Now, ∠RPX = ∠Q + ∠R ....[Exterior angle property]
⇒ 105° = x° + 2x°
⇒ 105° = 3x°
⇒ x = 35°
⇒ ∠Q = x° = 35° and ∠R = 2x° = 70°.
APPEARS IN
संबंधित प्रश्न
The exterior angles, obtained on producing the side of a triangle both ways, are 100° and 120°. Find all the angles of the triangle.
In a triangle PQR, ∠P + ∠Q = 130° and ∠P + ∠R = 120°. Calculate each angle of the triangle.
In the figure given below, if RS is parallel to PQ, then find the value of ∠y.
In a triangle PQR, the internal bisectors of angles Q and R meet at A and the external bisectors of the angles Q and R meet at B. Prove that: ∠QAR + ∠QBR = 180°.
Use the given figure to show that: ∠p + ∠q + ∠r = 360°.
In a triangle ABC, if the bisectors of angles ABC and ACB meet at M then prove that: ∠BMC = 90° + `(1)/(2)` ∠A.
If bisectors of angles A and D of a quadrilateral ABCD meet at 0, then show that ∠B + ∠C = 2 ∠AOD
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.
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.
