Question

plzzz answer this question only if u know the answer and also give the reasons for the answers and tell me how u solved the question. plz tell the correct answer only this answer is very important for me. plz zz*
InABC, BE and CF are altitudes and intersect at H. If m. angle BAC = 70 then m. angle BHC equals
(A) 105
(B) 110
(C) 115
(D) 120
​

Answer 1

☆Given :

• BAC = 70°

☆To find :

• BHC

note : both the altitudes are on the sides of triangle with 90° .

$\implies$ "H" Lies between ACF

• we know : lines along a straight path is equal to 180°.

$\therefore \: \angle$ H = 90 ° + 90°

= 180°

$\large \red{Now,}$

$\angle$ BHC = 90° (vertically opposite angles)

• Note : as the altitudes always have 90° where they end , they are interacting on point H .

$\therefore$ These sides "BH" and "CH" and their anhles are equal- (Or it is an isosceles triangle)

#Finding out :

$\rightarrow$ Angle sum property

$\implies$ 180° = 90° + B + C

$\implies$ 180° - 90° = B + C

$\implies$ 90° = B + C

• ( we know B and C angles are equal )

$\implies$ 90° ÷ 2 = B and C (each)

$\implies$ 45° = B and C (each)

• Angle BHC :

= BAC + angle B or C

= 70° + 45°

$\large \therefore \: \boxed{\angle h = 115 \degree}$