Advertisements
Advertisements
Question
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.
Solution
For ant triangle, sum of measures of all three angles = 180°
Thus, we have
( x + 10)° + ( x + 30)° + ( x - 10)° = 180°
⇒ x° + 10° + x° + 30° + x° - 10° = 180°
⇒ 3x° + 30° = 180°
⇒ 3x° = 150°
⇒ x = 50°
Now,
( x + 10)° = (50 + 10)° = 60°
(x + 30)° = (50 + 30)° = 80°
(x - 10)° = (50 - 10)° = 40°
Thus, the angles of a triangle are 60°, 80° and 40°.
APPEARS IN
RELATED QUESTIONS
In the given figure, ∠Q: ∠R = 1: 2. Find:
a. ∠Q
b. ∠R
In a triangle PQR, ∠P + ∠Q = 130° and ∠P + ∠R = 120°. Calculate each angle of the triangle.
Use the given figure to find the value of y in terms of p, q and r.
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°.
In a triangle ABC. If D is a point on BC such that ∠CAD = ∠B, then prove that: ∠ADC = ∠BAC.
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.
If the angles of a triangle are in the ratio 2: 4: 6; show that the triangle is a right-angled 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.