Exercise 15B
Page-186Question 1:
In the figure given alongside, find the measure of ∠ACD.
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Answer 1:
We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.
∠ACD= ∠CAB+∠CBA∠ACD=75°+45°=120°
∠ACD= ∠CAB+∠CBA∠ACD=75°+45°=120°
Question 2:
In the figure given alongside, find the values of x and y.
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Answer 2:
We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.
∴ ∠BAC +∠ABC =∠ACDx +68 = 130x = 62
Sum of the angles in any triangle is 180o.
∴ ∠BAC +∠ABC+∠ACB =180°62 + 68 + y = 180y = 50
∴ ∠BAC +∠ABC =∠ACDx +68 = 130x = 62
Sum of the angles in any triangle is 180o.
∴ ∠BAC +∠ABC+∠ACB =180°62 + 68 + y = 180y = 50
Question 3:
In the figure given alongside, find the values of x and y.
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Answer 3:
We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.
∴ ∠BAC + ∠CBA = ∠ACD32 + x = 65x = 33
Also, sum of the angles in any triangle is 180°.
∴ ∠BAC+∠CBA +∠ACB= 180°32 +33+ y = 180y = 115
∴ x= 33
y =115
∴ ∠BAC + ∠CBA = ∠ACD32 + x = 65x = 33
Also, sum of the angles in any triangle is 180°.
∴ ∠BAC+∠CBA +∠ACB= 180°32 +33+ y = 180y = 115
∴ x= 33
y =115
Question 4:
An exterior angle of a triangle measures 110° and its interior opposite angles are in the ratio 2 : 3. Find the angles of the triangle.
Answer 4:
Suppose the interior opposite angles are (2x)° and (3x)°.
We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.
∴ 3x +2x= 110
x = 22
The interior opposite angles are (2×22)° and (3×22)°, i.e. 44° and 66°.
Suppose the third angle of the triangle is y°.
Now, sum of the angles in any triangle is 180°.
∴ 44 + 66 + y = 180
y = 70
Hence, the angles of the triangle are 44°, 66° and 70°.
We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.
∴ 3x +2x= 110
x = 22
The interior opposite angles are (2×22)° and (3×22)°, i.e. 44° and 66°.
Suppose the third angle of the triangle is y°.
Now, sum of the angles in any triangle is 180°.
∴ 44 + 66 + y = 180
y = 70
Hence, the angles of the triangle are 44°, 66° and 70°.
Question 5:
An exterior angle of a triangle is 100° and its interior opposite angles are equal to each other. Find the measure of each angle of the triangle.
Answer 5:
Suppose the interior opposite angles of an exterior angle 100o are xo and xo.
We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.
∴ x + x = 100
2x= 100
x= 50
Also, sum of the angles of any triangle is 180°.
Let the measure of the third angle be y°.
∴ x + x + y = 180
50 + 50 + y= 180
y = 80
Hence, the angles are of the measures 50°, 50° and 80°.
We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.
∴ x + x = 100
2x= 100
x= 50
Also, sum of the angles of any triangle is 180°.
Let the measure of the third angle be y°.
∴ x + x + y = 180
50 + 50 + y= 180
y = 80
Hence, the angles are of the measures 50°, 50° and 80°.
Question 6:
In the figure given alongside, find:
(i) ∠ACD
(ii) ∠AED
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(i) ∠ACD
(ii) ∠AED
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Answer 6:
We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.
In △ABC:
∠ACD=∠BAC+∠ABC=25°+45°∠ACD=70°(ii) In △ECD:∠AED=∠ECD+∠EDC=70°+40°=>∠AED=110°
In △ABC:
∠ACD=∠BAC+∠ABC=25°+45°∠ACD=70°(ii) In △ECD:∠AED=∠ECD+∠EDC=70°+40°=>∠AED=110°
Question 7:
In the figure given alongside, find:
(i) ∠ACD
(ii) ∠ADC
(iii) ∠DAE
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(i) ∠ACD
(ii) ∠ADC
(iii) ∠DAE
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Answer 7:
Sum of the angles of a triangle is 180°.
In△ABC:∠BAC+∠CBA+∠ACB=180°∠BAC=180°-(40°+100°)=>∠BAC=40°
We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.
∠ACD=∠BAC+∠CBA=40°+40°=80°(i) ∠ACD=80°(ii) In △ ACD:∠CAD+∠ACD+∠ADC=180°=>∠ADC=180°-(50°+80°)=>∠ADC=50°∴ ∠ADC= 50°(iii) ∠DAB+∠DAE=180° (since BE is a straight line) ∠DAE=180°-(∠DAC+∠CAB)∠DAE=180°-(50°+40°)∠DAE=90°
In△ABC:∠BAC+∠CBA+∠ACB=180°∠BAC=180°-(40°+100°)=>∠BAC=40°
We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.
∠ACD=∠BAC+∠CBA=40°+40°=80°(i) ∠ACD=80°(ii) In △ ACD:∠CAD+∠ACD+∠ADC=180°=>∠ADC=180°-(50°+80°)=>∠ADC=50°∴ ∠ADC= 50°(iii) ∠DAB+∠DAE=180° (since BE is a straight line) ∠DAE=180°-(∠DAC+∠CAB)∠DAE=180°-(50°+40°)∠DAE=90°
Question 8:
In the figure given alongside, x : y = 2 : 3 and ∠ACD = 130°.
Find the values of x, y and z.
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Find the values of x, y and z.
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Answer 8:
xy=23⇒3x = 2y⇒x=23y
We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.
∴ ∠A + ∠B =∠ACD
x°+ y° = 130°
⇒2y3+y =130⇒5y =130×3 ⇒5y= 390⇒y = 78⇒x =23×78⇒x=52
Also, sum of the angles in any triangle is 180°
∴ x+ y + z = 180
z= 180- 78 - 52
z= 50
We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.
∴ ∠A + ∠B =∠ACD
x°+ y° = 130°
⇒2y3+y =130⇒5y =130×3 ⇒5y= 390⇒y = 78⇒x =23×78⇒x=52
Also, sum of the angles in any triangle is 180°
∴ x+ y + z = 180
z= 180- 78 - 52
z= 50
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